Example Question - differentiation
Here are examples of questions we've helped users solve.
Differentiation of xcosx and Probability of Drawing Specific Cards from a Deck
<p>The image displays two separate questions. I will provide the solutions for both.</p> <p>For the differentiation of \( x\cos{x} \) with respect to \( x \) using the first principle:</p> <p>We have \( f(x) = x\cos{x} \), we need to find \( f'(x) \) using the first principle:</p> <p>\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]</p> <p>\[ = \lim_{{h \to 0}} \frac{(x+h)\cos(x+h) - x\cos{x}}{h} \]</p> <p>We then expand and arrange the expression and apply the limit. However, without the options or further context for finding the value of \( k \), this part of the question is incomplete.</p> <p>For the probability question, assuming we are looking to find the probability of drawing 1 diamond and 3 spades:</p> <p>The total number of ways to draw 4 cards from a 52 card deck is \( C(52, 4) \).</p> <p>The number of ways to draw 1 diamond from the 13 available diamonds is \( C(13, 1) \).</p> <p>The number of ways to draw 3 spades from the 13 available spades is \( C(13, 3) \).</p> <p>The probability \( P \) of the event is:</p> <p>\[ P = \frac{C(13, 1) \cdot C(13, 3)}{C(52, 4)} \]</p> <p>\[ P = \frac{13 \cdot \frac{13!}{3!(13-3)!}}{\frac{52!}{4!(52-4)!}} \]</p> <p>We can then simplify the factorials to get the probability.</p>
Trigonometric and Logarithmic Equations
<p>\text{Given the equations:}</p> <p>[1]\: y = \sin^3{x} + \csc^5{x^3} + \tan^5{ \left( \sqrt{x^2 + 1^3} \right)}</p> <p>\text{Differentiate equation [1] with respect to } x:</p> <p>\frac{d}{dx}y = 3\sin^2{x}\cos{x} - 5\csc^6{x^3}(3x^2)\cot{x^3} + 5\tan^4{ \left( \sqrt{x^2 + 1^3} \right) }\frac{1}{2\sqrt{x^2 + 1^3}}(2x)(1+\tan^2{ \left( \sqrt{x^2 + 1^3} \right) })</p> <p>[2]\: y = x\cos(\ln{x}) + 10\sec{(\sqrt{x})}</p> <p>\text{Differentiate equation [2] with respect to } x:</p> <p>\frac{d}{dx}y = \cos(\ln{x}) - x\sin(\ln{x})\frac{1}{x} + 10\sec{(\sqrt{x})}\tan{(\sqrt{x})}\frac{1}{2\sqrt{x}}</p> <p>[3]\: y = \frac{\sin{(x^2-1)}}{x} + \cos{(\cos{x})^2} - 2\cot{(x^{-3})}</p> <p>\text{Differentiate equation [3] with respect to } x:</p> <p>\frac{d}{dx}y = \frac{x(2x)\cos{(x^2 - 1)} - \sin{(x^2 - 1)}}{x^2} - 2\cos{(\cos{x})}\sin{x}(-\sin{x}) + 2\csc^2{(x^{-3})}(3x^{-4})</p>
Calculate the Derivative of a Trigonometric Function
<p>\frac{dy}{dx} = \frac{d}{dx}(sin^3(x) + \csc^5(x) + \tan^5(\sqrt{x^2 + 1}))</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))\frac{d}{dx}(\sqrt{x^2 + 1})</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))\frac{1}{2\sqrt{x^2 + 1}}(2x)</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5x\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))(\frac{1}{\sqrt{x^2 + 1}})</p>
Differentiation of a Composite Function with Trigonometric Factor
Используем правило производной произведения $(uv)'=u'v+uv'$ и производной степенной функции и тригонометрической функции. <p>Дано: $f(x) = (4x^4 + 2)\cos(x)$.</p> <p>Найдем производную от $4x^4$: $(4x^4)' = 16x^3$.</p> <p>Производная от $\cos(x)$: $(\cos(x))' = -\sin(x)$.</p> <p>По правилу производной произведения:</p> <p>$f'(x) = (4x^4 + 2)' \cdot \cos(x) + (4x^4 + 2) \cdot (\cos(x))'$</p> <p>$f'(x) = (16x^3) \cdot \cos(x) + (4x^4 + 2) \cdot (-\sin(x))$</p> <p>$f'(x) = 16x^3\cos(x) - (4x^4 + 2)\sin(x)$</p>
Derivative of a Polynomial Expression
<p>Для того чтобы найти производную многочлена \( f(x) = 3x^2 + x^4 \), следует применить правило нахождения производной степенной функции: \((x^n)' = nx^{n-1}\).</p> <p>Тогда производная данного многочлена будет:</p> <p>\( f'(x) = (3x^2)' + (x^4)' \)</p> <p>\( f'(x) = 3 \cdot 2x^{2-1} + 4x^{4-1} \)</p> <p>\( f'(x) = 6x + 4x^3 \)</p>
Determining the Derivative of a Polynomial Function
<p>The derivative of \(y = 2t^4 - 10t^2 + 13t\) with respect to \(t\) is calculated by differentiating each term separately.</p> <p>The derivative of the first term \(2t^4\) is \(8t^3\).</p> <p>The derivative of the second term \(-10t^2\) is \(-20t\).</p> <p>The derivative of the third term \(13t\) is \(13\).</p> <p>Therefore, the derivative \( \frac{dy}{dt} = 8t^3 - 20t + 13 \).</p>
Derivative of a Polynomial Function
Given: y = 2t^4 - 10t^2 + 13t. To find: dy/dt. Solution: dy/dt = d/dt (2t^4) - d/dt (10t^2) + d/dt (13t) dy/dt = 2 * 4t^(4-1) - 10 * 2t^(2-1) + 13 * 1t^(1-1) dy/dt = 8t^3 - 20t + 13.
Derivative of a Polynomial
\[ \frac{dy}{dt} = \frac{d}{dt}(2t^4) - \frac{d}{dt}(10t^2) + \frac{d}{dt}(13t) \\ \frac{dy}{dt} = 8t^3 - 20t + 13 \]
Derivative of a polynomial function
\frac{dy}{dt} = \frac{d}{dt}(2t^4) - \frac{d}{dt}(10t^2) + \frac{d}{dt}(13t) \frac{dy}{dt} = 8t^3 - 20t + 13
Derivative of a Function with Rational Expression
Given the function: \( h(x) = \frac{4x^3 - 7x + 8}{x} \) First, simplify \( h(x) \) by dividing each term in the numerator by \( x \): \( h(x) = 4x^2 - 7 + \frac{8}{x} \) Now, find the derivative \( h'(x) \): \( h'(x) = \frac{d}{dx}(4x^2) - \frac{d}{dx}(7) + \frac{d}{dx}\left(\frac{8}{x}\right) \) \( h'(x) = 8x - 0 - 8x^{-2} \) \( h'(x) = 8x - 8x^{-2} \) Expressing \( -8x^{-2} \) as \( -\frac{8}{x^2} \), we have: \( h'(x) = 8x - \frac{8}{x^2} \)
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) \( \frac{2x}{x^2 + 1} \)
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: \( B) \) \( y' = \frac{3x + 17}{(2x + 5)^\frac{3}{2}} \)
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.
Derivatives of Mathematical Functions
The given image shows two mathematical functions, f(x) and g(x), and asks for the derivation of these functions at certain points. The functions are: f(x) = x^3 + 1 g(x) = x^2 - x You are asked to do two things: a) Find the derivative of f(x)g(x) at x = 1. b) Find the derivative of the quotient f(x)/g(x) at x = -1. Let's solve each part separately. a) To find the derivative of f(x)g(x), we first need to find the product of these functions and then take the derivative of the product, and evaluate it at x = 1. The product of f(x) and g(x) is: f(x)g(x) = (x^3 + 1)(x^2 - x) Instead of multiplying and then differentiating, we will use the product rule for differentiation, which states that (fg)' = f'g + fg'. Here are the derivatives of f(x) and g(x): f'(x) = d/dx (x^3 + 1) = 3x^2 g'(x) = d/dx (x^2 - x) = 2x - 1 Now apply the product rule: (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) = (3x^2)(x^2 - x) + (x^3 + 1)(2x - 1) Now we evaluate this derivative at x = 1: (3(1)^2)(1^2 - 1) + (1^3 + 1)(2(1) - 1) = 3(1)(0) + (2)(1) = 0 + 2 = 2 So the derivative of f(x)g(x) at x = 1 is 2. b) To find the derivative of the quotient f(x)/g(x), we will use the quotient rule for differentiation, which states that (f/g)' = (f'g - fg') / g^2. Again, we will need to use f'(x) = 3x^2 and g'(x) = 2x - 1, and f(x) = x^3 + 1 and g(x) = x^2 - x. Applying the quotient rule, we get: (f(x)/g(x))' = (f'(x)g(x) - f(x)g'(x)) / (g(x))^2 = ((3x^2)(x^2 - x) - (x^3 + 1)(2x - 1)) / (x^2 - x)^2 Now we evaluate this derivative at x = -1: = ((3(-1)^2)((-1)^2 - (-1)) - ((-1)^3 + 1)(2(-1) - 1)) / ((-1)^2 - (-1))^2 = ((3)(1)(1 + 1) - ((-1) + 1)(-2 - 1)) / (1 + 1)^2 = (6 - (0)(-3)) / 4 = 6 / 4 = 3 / 2 So the derivative of f(x)/g(x) at x = -1 is 3/2.
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 \(-1\).
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