Example Question - chain rule
Here are examples of questions we've helped users solve.
Finding the Derivative of Sine Function with a Scalar Multiple of the Variable
<p>\frac{d}{dx} [\sin(2x)]</p> <p>= \cos(2x) \cdot \frac{d}{dx}[2x]</p> <p>= \cos(2x) \cdot 2</p> <p>= 2\cos(2x)</p>
Finding Tangent Line to Composition of Functions
Given \( g(x) = \frac{16}{x} \) and \( f(x) = -4 \sqrt{x} \), find the tangent line to \( g(f(x)) \) at \( x=4 \). Step 1: Compute \( f(4) \). \( f(4) = -4 \sqrt{4} = -4 \cdot 2 = -8 \) Step 2: Compute \( g(f(4)) \). \( g(-8) = \frac{16}{-8} = -2 \) Step 3: Compute \( f'(x) \). \( f'(x) = -4 \cdot \frac{1}{2}x^{-\frac{1}{2}} = -2x^{-\frac{1}{2}} \) Step 4: Compute \( f'(4) \). \( f'(4) = -2 \cdot 4^{-\frac{1}{2}} = -2 \cdot \frac{1}{2} = -1 \) Step 5: Compute \( g'(x) \). \( g'(x) = -16x^{-2} \) Step 6: Compute \( g'(-8) \). \( g'(-8) = -16(-8)^{-2} = -16 \cdot \frac{1}{64} = -\frac{1}{4} \) Step 7: Apply chain rule to find derivative of \( g(f(x)) \) at \( x=4 \). \( (g(f(x)))' = g'(f(x)) \cdot f'(x) \) \( (g(f(4)))' = g'(-8) \cdot f'(4) = -\frac{1}{4} \cdot (-1) = \frac{1}{4} \) Step 8: Write the equation of the tangent line. \( y = mx + b \) \( m = \frac{1}{4} \) \( y = \frac{1}{4}x + b \) Step 9: Use point \( (f(4), g(f(4))) = (4, -2) \) to solve for \( b \). \( -2 = \frac{1}{4} \cdot 4 + b \) \( -2 = 1 + b \) \( b = -3 \) Step 10: Final equation of the tangent line. \( y = \frac{1}{4}x - 3 \)
Derivative of Inverse Cosine Function with Chain Rule
Para resolver la derivada de la función \( f(x) = \arccos\left( \frac{1}{\sqrt{x^2 + 1}} \right) \), utilizaremos la regla de la cadena. Primero, considera que la derivada de \( \arccos(u) \) es \( -\frac{1}{\sqrt{1 - u^2}} \), y luego debemos tomar la derivada de \( u \) respecto a \( x \), siendo \( u = \frac{1}{\sqrt{x^2 + 1}} \). La derivada de \( u \) con respecto a \( x \) es un poco más complicada, ya que tiene una raíz cuadrada en el denominador. Vamos a calcularla paso a paso. La función \( u \) puede ser reescrita como \( u(x) = (x^2 + 1)^{-1/2} \). Al diferenciar \( u \) con respecto a \( x \), aplicamos la regla de la cadena y obtenemos: \[ u'(x) = -\frac{1}{2} (x^2 + 1)^{-3/2} \cdot 2x = -\frac{x}{(x^2 + 1)^{3/2}} \] Ahora, aplicamos la regla de la cadena para diferenciar la función \( f(x) \): \[ \begin{aligned} f'(x) &= -\frac{1}{\sqrt{1 - u^2}} \cdot u'(x) \\ &= -\frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{x^2 + 1}}\right)^2}} \cdot \left(-\frac{x}{(x^2 + 1)^{3/2}}\right) \\ &= -\frac{1}{\sqrt{1 - \frac{1}{x^2 + 1}}} \cdot \left(-\frac{x}{(x^2 + 1)^{3/2}}\right) \\ &= -\frac{1}{\sqrt{\frac{x^2 + 1}{x^2 + 1} - \frac{1}{x^2 + 1}}} \cdot \left(-\frac{x}{(x^2 + 1)^{3/2}}\right) \\ &= -\frac{1}{\sqrt{\frac{x^2}{x^2 + 1}}} \cdot \left(-\frac{x}{(x^2 + 1)^{3/2}}\right) \\ &= -\frac{1}{\frac{x}{\sqrt{x^2 + 1}}} \cdot \left(-\frac{x}{(x^2 + 1)^{3/2}}\right) \\ &= \frac{x}{(x^2 + 1)^{2}}. \end{aligned} \] Entonces, la respuesta correcta es: a) \( \frac{2x}{x^2 + 1} \)
Partial Derivative Quotient Calculation of Ln(xy)
好的，题目要求我们求函数\( z = \ln(xy) \)的偏导数的商 \(\frac{\partial^3 z}{\partial x^2 \partial y} \div \frac{\partial^3 z}{\partial x \partial y^2}\) 的值。 首先，我们计算 \( z = \ln(xy) \) 的一阶和二阶偏导数： 一阶偏导数： \(\frac{\partial z}{\partial x} = \frac{1}{xy} \cdot y = \frac{1}{x}\) \(\frac{\partial z}{\partial y} = \frac{1}{xy} \cdot x = \frac{1}{y}\) 二阶偏导数： \(\frac{\partial^2 z}{\partial x^2} = -\frac{1}{x^2}\) \(\frac{\partial^2 z}{\partial y^2} = -\frac{1}{y^2}\) \(\frac{\partial^2 z}{\partial x \partial y} = -\frac{1}{xy^2}\) 三阶偏导数： \(\frac{\partial^3 z}{\partial x^2 \partial y} = \frac{2}{xy^3}\) \(\frac{\partial^3 z}{\partial x \partial y^2} = \frac{2}{x^2 y^2}\) 现在我们可以计算商的值了： \[ \frac{\frac{\partial^3 z}{\partial x^2 \partial y}}{\frac{\partial^3 z}{\partial x \partial y^2}} = \frac{\frac{2}{xy^3}}{\frac{2}{x^2 y^2}} = \frac{2}{xy^3} \cdot \frac{x^2 y^2}{2} = \frac{x}{y} \] 所以，该导数的商就是 \(\frac{x}{y}\)。
Derivative of a Quotient Function
Để tìm đảo hàm của hàm số \( y = \frac{3x - 2}{\sqrt{2x + 5}} \), ta sử dụng quy tắc đạo hàm của thương. Nếu ta có \( u(x) \) và \( v(x) \) là hai hàm số khả vi và \( v(x) \neq 0 \), thì đạo hàm của thương là: \[ (u/v)' = \frac{u'v - uv'}{v^2} \] Ở đây, ta thiết lập \( u(x) = 3x - 2 \) và \( v(x) = \sqrt{2x + 5} \). Ta cần tìm \( u'(x) \) và \( v'(x) \) trước. Đạo hàm của \( u(x) \) là: \[ u'(x) = (3x - 2)' = 3 \] Để tìm đạo hàm của \( v(x) \), vì \( v(x) = (2x + 5)^{1/2} \), ta sử dụng quy tắc chuỗi để tìm đạo hàm của hàm số mũ: \[ v'(x) = \frac{1}{2}(2x + 5)^{-1/2} \cdot (2x + 5)' = \frac{1}{2}(2x + 5)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x + 5}} \] Bây giờ ta có thể tìm đạo hàm của thương: \[ y' = \frac{u'v - uv'}{v^2} = \frac{3\sqrt{2x + 5} - (3x - 2)\frac{1}{\sqrt{2x + 5}}}{2x + 5} \] \[ y' = \frac{3(2x + 5) - (3x - 2)}{(2x + 5)^\frac{3}{2}} \] \[ y' = \frac{6x + 15 - 3x + 2}{(2x + 5)^\frac{3}{2}} \] \[ y' = \frac{3x + 17}{(2x + 5)^\frac{3}{2}} \] Nhìn vào các phương án, ta thấy đáp án đúng là: \( B) \) \( y' = \frac{3x + 17}{(2x + 5)^\frac{3}{2}} \)
Calculating Differentials for a Given Function
La pregunta está pidiendo calcular el diferencial \( dy \) y el diferencial \( d_{y} \) para la función \( f(x)=e^{2ln(x^{2} + 1)} \sin{(x^{2} - 2y)} \) dado que \( \Delta x = 0,02 \). Para encontrar \( dy \), necesitamos calcular la derivada parcial de \( f \) con respecto a \( x \) y luego usarla para multiplicarla por el cambio en \( x \), que es \( \Delta x \). Para encontrar \( df \), necesitamos calcular el gradiente de \( f \), es decir, necesitamos encontrar tanto la derivada parcial con respecto a \( x \) como la derivada parcial con respecto a \( y \), y después usarlas para multiplicar por los diferenciales correspondientes \( dx \) y \( dy \). Voy a proceder a calcular \( \frac{\partial f}{\partial x} \) y \( \frac{\partial f}{\partial y} \). Primero, calculamos la derivada parcial de \( f \) con respecto a \( x \): \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( e^{2ln(x^{2} + 1)} \sin{(x^{2} - 2y)} \right) \] Para esto, primero notamos que \( e^{2ln(x^{2} + 1)} = (x^{2} + 1)^{2} \), simplificando el cálculo. \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( (x^{2} + 1)^{2} \sin{(x^{2} - 2y)} \right) \] Usamos la regla de la cadena y la regla del producto para esta derivada: \[ \frac{\partial f}{\partial x} = 2(x^{2} + 1) \cdot 2x \sin{(x^{2} - 2y)} + (x^{2} + 1)^{2} \cos{(x^{2} - 2y)} \cdot 2x \] \[ \frac{\partial f}{\partial x} = 4x(x^{2} + 1) \sin{(x^{2} - 2y)} + 2x(x^{2} + 1)^{2} \cos{(x^{2} - 2y)} \] Ahora calculamos la derivada parcial de \( f \) con respecto a \( y \): \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( (x^{2} + 1)^{2} \sin{(x^{2} - 2y)} \right) \] Necesitamos usar la regla de la cadena para esta derivada: \[ \frac{\partial f}{\partial y} = -2(x^{2} + 1)^{2} \cos{(x^{2} - 2y)} \] Con estas derivadas encontradas, ahora podemos calcular \( dy \) y \( df \): Para \( dy \), usando \( \Delta x \) en lugar \( dx \), ya que no se proporciona un valor para \( dy \), tendríamos algo así: \[ dy = \frac{\partial f}{\partial x} \cdot \Delta x \] \[ dy = \left( 4x(x^{2} + 1) \sin{(x^{2} - 2y)} + 2x(x^{2} + 1)^{2} \cos{(x^{2} - 2y)} \right) \cdot 0,02 \] Para \( df \) (también notado como \( d_{y} \) en la pregunta, pero esto parece ser un error tipográfico, ya que \( d_{y} \) normalmente representaría una derivada parcial con respecto a \( y \), no un diferencial completo), se calcularía de la siguiente manera: \[ df = \frac{\partial f}{\partial x} \cdot dx + \frac{\partial f}{\partial y} \cdot dy \] \[ df = \left( 4x(x^{2} + 1) \sin{(x^{2} - 2y)} + 2x(x^{2} + 1)^{2} \cos{(x^{2} - 2y)} \right) \cdot dx - 2(x^{2} + 1)^{2} \cos{(x^{2} - 2y)} \cdot dy \] Sin valores específicos para \( x \), \( y \), \( dx \), y \( dy \) (excepto \( \Delta x \)), no podemos dar un valor numérico a \( dy \) y \( df \). Debes reemplazar \( dx \) con \( \Delta x \) y usar los valores de \( x \) y \( y \) dados en el problema para encontrar los valores numéricos de \( dy \) y \( df \).
Derivative of a Function Involving Square Roots and Chains
The function is \( f(x) = x \sqrt{1 + \sqrt{2x + 6}} \). To find \( f'(x) \), we'll need to use the product rule and the chain rule. The product rule states that if you have a function \( u(x) \cdot v(x) \), then the derivative of this function with respect to \( x \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). The chain rule states that the derivative of a composite function \( u(v(x)) \) is \( u'(v(x)) \cdot v'(x) \). Let \( u(x) = x \) and \( v(x) = \sqrt{1 + \sqrt{2x + 6}} \). Then \( u'(x) = 1 \) since the derivative of \( x \) with respect to \( x \) is 1. For \( v(x) \), let \( h(x) = 1 + \sqrt{2x + 6} \), so \( v(x) = \sqrt{h(x)} \). First, find the derivative of \( h(x) \): \( h'(x) = 0 + \frac{1}{2} (2x + 6)^{-\frac{1}{2}} \cdot (2) = \frac{1}{\sqrt{2x + 6}} \). Next, we use the chain rule to find the derivative of \( v(x) \): \( v'(x) = \frac{1}{2} h(x)^{-\frac{1}{2}} \cdot h'(x) = \frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} \). Now we can find \( f'(x) \) using the product rule: \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \) \( f'(x) = 1 \cdot \sqrt{1 + \sqrt{2x + 6}} + x \cdot \left(\frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} \right) \) \( f'(x) = \sqrt{1 + \sqrt{2x + 6}} + \frac{x}{2(1 + \sqrt{2x + 6})\sqrt{2x + 6}} \). This is the derivative of the given function. Be sure to simplify where possible to get the final expression for \( f'(x) \).
Differentiation of Mathematical Functions
The image is blurry, but it seems to show four mathematical functions that need to be differentiated with respect to \( x \). I'll attempt to identify them and differentiate each one: (a) \( y = (x + n)^1 \) Using the power rule, we get: \( \frac{dy}{dx} = 1 \cdot (x + n)^{1 - 1} \cdot \frac{d}{dx}(x + n) \) \( \frac{dy}{dx} = (x + n)^0 \cdot \frac{d}{dx}(x + n) \) Since \( (x + n)^0 = 1 \) and \( \frac{d}{dx}(x + n) = 1 \), then \( \frac{dy}{dx} = 1 \cdot 1 = 1 \) (b) \( f(x) = (2x + n)^1 \) Applying the power rule: \( f'(x) = 1 \cdot (2x + n)^{1 - 1} \cdot \frac{d}{dx}(2x + n) \) \( f'(x) = (2x + n)^0 \cdot \frac{d}{dx}(2x + n) \) Since \( (2x + n)^0 = 1 \) and \( \frac{d}{dx}(2x + n) = 2 \), then \( f'(x) = 1 \cdot 2 = 2 \) (c) \( y = (3 - 4x)^5 \) Using the chain rule, we get: \( \frac{dy}{dx} = 5(3 - 4x)^{4}(-4) \) \( \frac{dy}{dx} = -20(3 - 4x)^{4} \) (d) \( g(x) = (3z - 4x)^2 \) (There appears to be a typo in the variable used in the original function. Assuming it's supposed to be \( x \), not \( z \), and differentiating accordingly:) \( g'(x) = 2(3 - 4x)^1(-4) \) \( g'(x) = -8(3 - 4x) \) If the variables \( n \) or \( z \) are constants, then my differentiation is correct. If \( n \) or \( z \) are not constants, and you meant a different variable or power, please provide the correct expressions.
Derivative of a Function with Respect to x using Quotient Rule and Chain Rule
The image shows a mathematical problem asking to find the derivative with respect to \(x\) of the function \(f(x,y)\), which is provided as: \[ \frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right) \] To find this derivative, we will use the quotient rule along with the chain rule because the function has both \(x\) and \(y\) variables, where \(y\) could be a function of \(x\). The quotient rule is: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \] where \(u = 2xy\) and \(v = \sqrt{x^2 + y^2}\). The derivative of \(u = 2xy\) with respect to \(x\) is \(u' = 2y + 2xy'\) (using the product rule and assuming \(y\) is a function of \(x\)). The derivative of \(v = \sqrt{x^2 + y^2}\) with respect to \(x\) is \(v' = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x + 2yy')\) (using the chain rule and assuming \(y\) is a function of \(x\)). Simplifying gives \(v' = \frac{x + yy'}{\sqrt{x^2 + y^2}}\). Now, apply the quotient rule: \[ \frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right) = \frac{(2y + 2xy')\sqrt{x^2 + y^2} - 2xy\left(\frac{x + yy'}{\sqrt{x^2 + y^2}}\right)}{x^2 + y^2} \] Simplify the expression: \[ = \frac{2y(x^2 + y^2) + 2xy'y\sqrt{x^2 + y^2} - 2x^2y - 2xy^2y'}{(x^2 + y^2)^{3/2}} \] Now, combine like terms and simplify further: \[ = \frac{2y^3 + 2xy'y\sqrt{x^2 + y^2} - 2xy^2y'}{(x^2 + y^2)^{3/2}} \] If \(y\) is not a function of \(x\), \(y'\) would simply be 0, and the expression would be further simplified. However, the derivative expressed as it stands takes into consideration that \(y\) may be a function of \(x\). If you have more specific information about \(y\), such as whether it is a constant or a function of \(x\), you can simplify the expression accordingly.
Finding Second Derivative of Natural Logarithm Function
The image shows a mathematical expression and a request to find the second derivative of the function with respect to x. The function is \( y = \ln(2x^2 + x) \). We need to find \( \frac{d^2y}{dx^2} \) when \( x = -1 \). First, let's find the first derivative \( \frac{dy}{dx} \). To do this, we'll use the chain rule since y is the natural logarithm of a function of x. Let \( u = 2x^2 + x \), so \( y = \ln(u) \). First, let's differentiate \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d(2x^2 + x)}{dx} = 4x + 1 \] Now let's differentiate \( y \) with respect to \( u \): \[ \frac{dy}{du} = \frac{d(\ln(u))}{du} = \frac{1}{u} \] Now use the chain rule to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot (4x + 1) = \frac{4x + 1}{2x^2 + x} \] Next, we'll find the second derivative \( \frac{d^2y}{dx^2} \). For this, we need to differentiate \( \frac{dy}{dx} \) with respect to \( x \) using the quotient rule: Let \( v = 4x + 1 \) and \( w = 2x^2 + x \), so \( v' = 4 \) and \( w' = 4x + 1 \). Now apply the quotient rule \( \left( \frac{v}{w} \right)' = \frac{v'w - vw'}{w^2} \): \[ \frac{d^2y}{dx^2} = \frac{v'w - vw'}{w^2} = \frac{4(2x^2 + x) - (4x + 1)(4x + 1)}{(2x^2 + x)^2} \] Now simplify and find \( \frac{d^2y}{dx^2} \) when \( x = -1 \): \[ \frac{d^2y}{dx^2} = \frac{8x^2 + 4x - (16x^2 + 8x + 1)}{(2x^2 + x)^2} \] \[ \frac{d^2y}{dx^2} = \frac{8x^2 + 4x - 16x^2 - 8x - 1}{(2x^2 + x)^2} \] \[ \frac{d^2y}{dx^2} = \frac{-8x^2 - 4x - 1}{(2x^2 + x)^2} \] Now plug in \( x = -1 \): \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-8(-1)^2 - 4(-1) - 1}{(2(-1)^2 - 1)^2} \] \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-8 + 4 - 1}{(2 + (-1))^2} \] \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-5}{1^2} \] \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = -5 \] The second derivative of the function at \( x = -1 \) is \( -5 \).
Derivatives in Calculus
The image shows three mathematical problems labeled a), b), and c), all of which are related to calculus. I'll assist you with each one separately. a) Given \( f(x) = 2x^2 - 5x + 1 \), you are asked to find \( f'(x) \), which is the first derivative of the function with respect to x. To find \( f'(x) \), differentiate \( f(x) \) term by term: \[ f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(5x) + \frac{d}{dx}(1) \] Using the power rule \( \frac{d}{dx}x^n = nx^{n-1} \) and the constant rule \( \frac{d}{dx}c = 0 \) for a constant c, we get: \[ f'(x) = 2 \cdot 2x^{2-1} - 5 \cdot 1x^{1-1} \] Simplify the expression: \[ f'(x) = 4x - 5 \] So, the first derivative \( f'(x) \) of the function \( f(x) = 2x^2 - 5x + 1 \) is \( f'(x) = 4x - 5 \). b) Next, you are asked to find the derivative \( \frac{dy}{dx} \) given \( y = 3(x^2 - x)^2 \). We'll use the chain rule here: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), where \( u = x^2 - x \), hence \( y = 3u^2 \). First, we differentiate \( y \) with respect to \( u \): \[ \frac{dy}{du} = 3 \cdot 2u^{2-1} = 6u \] Then, we differentiate \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(x) = 2x - 1 \] Now, multiply both derivatives together to get \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 6u(2x - 1) \] Substitute \( u \) back with \( x^2 - x \): \[ \frac{dy}{dx} = 6(x^2 - x)(2x - 1) \] So, \( \frac{dy}{dx} = 6(x^2 - x)(2x - 1) \) is the derivative of \( y = 3(x^2 - x)^2 \) with respect to \( x \). c) Finally, you are given \( f(x) = ax^2 + 3x - 1 \) and asked to find the value of \( a \), given that the limit \( \lim_{x \to -3} \frac{f(x) - f(-3)}{x - (-3)} = 13 \). This limit represents the derivative of \( f(x) \) at \( x = -3 \). Let's first find the derivative \( f'(x) \): \[ f'(x) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(1) \] \[ f'(x) = 2ax + 3 \] Now, evaluate \( f'(x) \) at \( x = -3 \): \[ f'(-3) = 2a(-3) + 3 = -6a + 3 \] We are given that \( f'(-3) = 13 \), so: \[ -6a + 3 = 13 \] Solving for \( a \): \[ -6a = 13 - 3 \] \[ -6a = 10 \] \[ a = -\frac{10}{6} \] \[ a = -\frac{5}{3} \] Therefore, \( a = -\frac{5}{3} \).
Finding the Derivative of a Function Expression
The image shows two graphs, y = f(x) and y = g(x), intersecting on a Cartesian coordinate plane, and there are lines \( d_1 \) and \( d_2 \) which appear to be parallel to the x-axis at different y-values. The function \( h(x) \) is defined such that \( h(x + 1) = f(x) - g(x + 2) \). We are tasked with finding the value of \( h'(2) \). The first step to find \( h'(2) \) is to differentiate the equation for \( h(x + 1) \) with respect to \( x \). So, let's differentiate \( h(x + 1) \) using the chain rule: If \( u = x + 1 \), then \( du/dx = 1 \) and so, \[ h'(u) = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{dh}{du} \] Now differentiate \( f(x) \) and \( g(x + 2) \) separately with respect to \( x \). Using the property \( \frac{d}{dx}f(x) = f'(x) \) and \( \frac{d}{dx}g(x+2) \) with the chain rule (\( v = x + 2 \), \( dv/dx = 1 \)), we get: \[ h'(u) = f'(x) - g'(v) \] Knowing that \( h'(u) = h'(x + 1) \), we can convert the differentiation point from \( x \) to \( u \) to find \( h'(2) \): \[ h'(2) = f'(1) - g'(4) \] To get the values of \( f'(1) \) and \( g'(4) \), we need to find the slopes of \( f(x) \) and \( g(x) \) at \( x = 1 \) and \( x = 4 \), respectively. Looking at the graph, \( f(x) \) passes through point A(1,0), and since \( d_1 \) and \( d_2 \) are parallel to the x-axis, it means the slope is 0 at \( x = 1 \) for \( f(x) \). Therefore, the slope \( f'(1) = 0 \). The graph doesn't give us \( g(4) \) directly, but we can use the fact that the line through the point B(-1, 2) is at a 45-degree angle to infer that this line, representing \( g(x) \), has a slope of 1, as a 45-degree line in the first quadrant will have a slope of 1. Hence, \( g'(x) = 1 \) for all \( x \) in the domain of \( g \), and specifically \( g'(4) = 1 \). Now, substitute the slope values into the equation for \( h'(2) \): \[ h'(2) = 0 - 1 = -1 \] Thus, the value of \( h'(2) \) is \(-1\).
Calculus - Derivative of y = 3 cos(2x)
The image shows a calculus problem asking to find the derivative dy/dx of the function y with respect to x. The function given is: y = 3 cos(2x) To find the derivative of this function, we use the chain rule, which in this case involves differentiating the cosine function and then accounting for the inner function, 2x. The derivative of cos(u) with respect to u is -sin(u), and then we multiply this by the derivative of the inner function, which is 2 in this case. The derivative of y with respect to x is thus: dy/dx = -3 sin(2x) * 2 Simplify the expression: dy/dx = -6 sin(2x)
Derivative of Composite Trigonometric Function
To find the derivative of the function \( y = \cos(e^{3\theta^2}) \), we will apply the chain rule, which says that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Let \( u = e^{3\theta^2} \), so our function becomes \( y = \cos(u) \). Taking the derivative with respect to \(\theta\), we get: \( \frac{dy}{du} = -\sin(u) \) (derivative of the outer function, \( \cos(u) \) with respect to \( u \)) Now, we find \( \frac{du}{d\theta} \): \( u = e^{3\theta^2} \) implies \( \frac{du}{d\theta} = \frac{d}{d\theta} e^{3\theta^2} \) To find this, we use the chain rule again: Let \( v = 3\theta^2 \), then \( u = e^v \) Take the derivative of \( v \) with respect to \( \theta \): \( \frac{dv}{d\theta} = 6\theta \) Now, take the derivative of \( u \) with respect to \( v \): \( \frac{du}{dv} = e^v \) So the derivative of \( u \) with respect to \( \theta \) is: \( \frac{du}{d\theta} = \frac{du}{dv} \cdot \frac{dv}{d\theta} = e^v \cdot 6\theta = e^{3\theta^2} \cdot 6\theta \) Now, we use the chain rule to find the derivative of \( y \) with respect to \( \theta \): \( \frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta} = -\sin(u) \cdot e^{3\theta^2} \cdot 6\theta \) Substitute \( u \) back in to get: \( \frac{dy}{d\theta} = -\sin(e^{3\theta^2}) \cdot e^{3\theta^2} \cdot 6\theta \) So the answer is: \[ \frac{dy}{d\theta} = -6\theta e^{3\theta^2} \sin(e^{3\theta^2}) \]
Derivative Calculation Using Logarithmic Differentiation
To differentiate the given function \( f(x) = \frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7} \) using logarithmic differentiation, we first take the natural logarithm of both sides of the equation defining \( f(x) \): 1. Take the natural logarithm of f(x): \[ \ln(f(x)) = \ln\left(\frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7}\right) \] 2. Apply properties of logarithms to simplify the right-hand side: \[ \ln(f(x)) = \ln(x^3) + \ln((3 - 2x)^2) - \ln(\sqrt[3]{x} + 7) \] \[ \ln(f(x)) = 3\ln(x) + 2\ln(3 - 2x) - \ln(\sqrt[3]{x} + 7) \] 3. Differentiate both sides with respect to x: \[ \frac{1}{f(x)} \cdot f'(x) = \frac{d}{dx}(3\ln(x)) + \frac{d}{dx}(2\ln(3 - 2x)) - \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) \] 4. Apply the derivatives; - For \( \frac{d}{dx}(3\ln(x)) \), use the derivative of \(\ln(x)\), which is \( \frac{1}{x} \): \[ \frac{d}{dx}(3\ln(x)) = 3 \cdot \frac{1}{x} = \frac{3}{x} \] - For \( \frac{d}{dx}(2\ln(3 - 2x)) \), use the chain rule: \[ \frac{d}{dx}(2\ln(3 - 2x)) = 2 \cdot \frac{1}{(3 - 2x)} \cdot (-2) = \frac{-4}{(3 - 2x)} \] - For \( \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) \), again use the chain rule: \[ \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) = \frac{1}{(\sqrt[3]{x} + 7)} \cdot \frac{d}{dx}(\sqrt[3]{x}) \] \[ \frac{d}{dx}(\sqrt[3]{x}) = \frac{1}{3x^{2/3}} \] So, \[ \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) = \frac{1}{(\sqrt[3]{x} + 7)} \cdot \frac{1}{3x^{2/3}} \] 5. Now combine these results: \[ \frac{1}{f(x)} \cdot f'(x) = \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \] 6. Solve for \( f'(x) \): \[ f'(x) = f(x) \cdot \left( \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \right) \] 7. Replace \( f(x) \) with the original function: \[ f'(x) = \frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7} \cdot \left( \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \right) \] This is the derivative of the function using logarithmic differentiation. You can simplify the expression further by combining terms and multiplying through by the original function.
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