Example Question - derivative calculation
Here are examples of questions we've helped users solve.
Maclaurin Series Derivative Calculation
很抱歉，图片中的内容看起来模糊不清，这导致我无法准确辨识图片中的所有文字。然而，根据你的文字描述和图片中可见的内容，我可以猜测这是一个关于Maclaurin级数的问题。 给出函数 \( g(t) = e^{t^3} \) ，我们需要利用 Maclaurin 级数求 \( g^{(21)}(0) \)，即函数 \( g(t) = e^{t^3} \) 在 \( t=0 \) 时的第21阶导数。 Maclaurin 级数展开公式为： \[ g(t) = g(0) + g'(0)t + \frac{g''(0)t^2}{2!} + \frac{g'''(0)t^3}{3!} + \cdots + \frac{g^{(n)}(0)t^n}{n!} + \cdots \] 我们知道 \( e^{t^3} \) 的 Maclaurin 级数是所有项的和 \( \sum_{n=0}^{\infty} \frac{t^{3n}}{n!} \) 因为 \(e^x \) 的 Maclaurin 级数是 \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \)。 在 \( t^3 \) 的情况下，这变为 \( e^{t^3} = 1 + t^3 + \frac{t^6}{2!} + \frac{t^9}{3!} + \frac{t^{12}}{4!} + \cdots \)。 我们寻找 \( t=0 \) 时函数的第21阶导数，相当于找系数 \( t^{21} \) 项的系数，并乘以 \( 21! \) 。 在上述展开中，\( t^{21} \) 的系数必定来自 \( \frac{t^{3n}}{n!} \)，其中 \( 3n = 21 \)，解得 \( n = 7 \)。 所以，\( t^{21} \) 的系数是 \( \frac{1}{7!} \)，并且为了得到 \( g^{(21)}(0) \)，我们需要将这个系数乘以 \( 21! \)。 计算可得： \[ g^{(21)}(0) = \frac{21!}{7!} \] 但是由于我们是在寻找 Maclaurin 级数的系数，并非真正计算导数，因此实际上 \( g^{(21)}(0) \) 就是直接的系数，不需要再乘 \( 21! \)。 所以 \( g^{(21)}(0) = \frac{1}{7!} \)。 选择项 (B) \( \frac{1}{7!} \) 是正确答案。
Derivative Calculation of a Function
La imagen muestra una pregunta de cálculo sobre derivadas. La función dada es \( f(x) = x^{2/3} \), y se nos pide determinar \( f'(x) \), es decir, la derivada primera de \( f(x) \). Para resolver este problema, utilizaremos la regla de potencias para la derivación. La regla dice que la derivada de \( x^n \) es \( n \cdot x^{n-1} \). Veamos cómo aplicar esto a nuestra función. Tenemos \( f(x) = x^{2/3} \). Entonces, utilizando la regla de potencias: \( f'(x) = \frac{2}{3} \cdot x^{\frac{2}{3} - 1} \) \( f'(x) = \frac{2}{3} \cdot x^{-\frac{1}{3}} \) \( f'(x) = \frac{2}{3} \cdot \frac{1}{x^{\frac{1}{3}}} \) \( f'(x) = \frac{2}{3x^{\frac{1}{3}}} \) Por lo tanto, la derivada de \( f(x) = x^{2/3} \) es \( f'(x) = \frac{2}{3x^{1/3}} \), que en las opciones correspondientes a la pregunta sería la opción B.
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 \( -5 \).
Derivative of a Composition of Functions using the Chain Rule
The given function is \( y = 3(x^2 - x)^3 \). To find \( \frac{dy}{dx} \), we need to differentiate this function with respect to \( x \). This is a composition of functions, and therefore we will need to apply the chain rule to differentiate it. The chain rule states that if we have a function \( h(x) = f(g(x)) \), then its derivative is \( h'(x) = f'(g(x)) \cdot g'(x) \). In your case, let \( u = x^2 - x \). Then the function \( y \) can be rewritten as \( y = 3u^3 \). Now, we will differentiate both sides with respect to \( x \): 1. Differentiate \( y \) with respect to \( u \): \[ \frac{dy}{du} = 3 \cdot 3u^2 = 9u^2 \] 2. Differentiate \( u = x^2 - x \) with respect to \( x \): \[ \frac{du}{dx} = 2x - 1 \] Now, apply the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 9u^2(2x - 1) \] Substitute \( u \) back in terms of \( x \): \[ \frac{dy}{dx} = 9(x^2 - x)^2(2x - 1) \] Remember to expand and simplify the expression if needed. But as per the instruction given in the question, you wanted the expression for \( \frac{dy}{dx} \), which is what we've found.
Calculating Derivative of a Function and Finding its Value
The image shows a function f(x) = 2x^7 - 5x + 1 and asks to calculate f'(x) given f(x) = 1. To find f'(x), we need to differentiate f(x) with respect to x. Taking the derivative term by term, we get: f'(x) = d/dx (2x^7) - d/dx (5x) + d/dx (1) Using the power rule for differentiation, which states that d/dx (x^n) = n*x^(n-1), we get: f'(x) = 7 * 2x^(7-1) - 5 * 1x^(1-1) f'(x) = 14x^6 - 5 Now we have an expression for f'(x). However, the problem states that this should be evaluated given f(x) = 1. This means we have to find x such that 2x^7 - 5x + 1 = 1. From the equation, we can simplify to: 2x^7 - 5x = 0 Removing the constant term from both sides since it does not impact finding x for which f(x) = 1, we need an additional step to find x that satisfies this equation. Generally, this would require solving a 7th-degree polynomial, which may or may not be solvable in terms of radicals. However, the task does not specify needing to find the value of x that makes f(x) = 1. Instead, if you need to find f'(1), then you would substitute x with 1: f'(1) = 14(1)^6 - 5 f'(1) = 14 - 5 f'(1) = 9 So, if you're required to find f'(1), the answer is 9. If you have to find f'(x) for the value of x such that f(x) = 1, the determination of x from the given equation 2x^7 - 5x + 1 = 1 will be more complex and may not have a straightforward solution.
Derivative Calculation with a Function and Point on a Graph
The image includes a graph displaying a straight line representing a function y = f(x), along with a pair of points labeled A(2, 1). Additionally, there's a mathematical question that states: Assuming the function g(x + 1) = (x^2 + 1) - f(x), what is the value of g'(3)? First, let's determine the slope of the line representing y = f(x) since we have the point A(2, 1) on the line. Since the line passes through the point A(2, 1) and the origin (0, 0), we can use these two points to calculate the slope (m): m = (y2 - y1) / (x2 - x1) m = (1 - 0) / (2 - 0) m = 1 / 2 Thus, the slope of the line is 1/2, and knowing that a line equation with slope m passing through the origin (0,0) has the form y = mx, the equation for f(x) is: f(x) = (1/2)x Now, let's put f(x) into the equation for g(x + 1): g(x + 1) = (x^2 + 1) - (1/2)x To find g'(x), the derivative of g(x), we must first write g(x) in terms of x: g(x) = ((x - 1)^2 + 1) - (1/2)(x - 1) g(x) = (x^2 - 2x + 1 + 1) - (1/2)x + (1/2) g(x) = x^2 - (5/2)x + 2 The next step is to differentiate g(x) with respect to x to find g'(x): g'(x) = 2x - 5/2 Finally, to find g'(3), substitute 3 into the derivative: g'(3) = 2(3) - 5/2 g'(3) = 6 - 5/2 g'(3) = (12/2) - (5/2) g'(3) = (12 - 5)/2 g'(3) = 7/2 Therefore, g'(3) is 7/2.
Derivative Calculation for a Function
The function f(x) provided in the image is: f(x) = (3x^5 - x^5) / 18x To differentiate this function, let's first simplify the expression f(x) by combining like terms in the numerator: f(x) = (3x^5 - x^5) / 18x = (2x^5) / 18x = (1/9)x^4 Now that we have the simplified form of f(x), we can differentiate it with respect to x: f'(x) = d/dx [(1/9)x^4] = (1/9) * d/dx [x^4] = (1/9) * (4x^3) Therefore, f'(x) = (4/9)x^3 Hence, the derivative of the function f(x) is (4/9)x^3 and is differentiable everywhere except at x = 0, since the original function included a term in the denominator with x, which would not be defined for x = 0.
Derivative Calculation using Logarithmic Differentiation
To find the derivative f'(1) of the function f(x) using logarithmic differentiation, we start by taking the natural logarithm of both sides of the function. Given: f(x) = (2x - 1)^15 * (5x - 4)^5 * (9x - 8)^5 / (11x - 10)^2 * (14x - 13)^3 * (17x - 16)^3 Let y = f(x), then ln(y) = ln(f(x)) Taking logarithms: ln(y) = 15 * ln(2x - 1) + 5 * ln(5x - 4) + 5 * ln(9x - 8) - 2 * ln(11x - 10) - 3 * ln(14x - 13) - 3 * ln(17x - 16) Now, differentiate both sides with respect to x: 1/y * dy/dx = 15/(2x - 1) * 2 + 5/(5x - 4) * 5 + 5/(9x - 8) * 9 - 2/(11x - 10) * 11 - 3/(14x - 13) * 14 - 3/(17x - 16) * 17 Simplify: dy/dx = y * [ 30/(2x - 1) + 25/(5x - 4) + 45/(9x - 8) - 22/(11x - 10) - 42/(14x - 13) - 51/(17x - 16) ] Now we need to substitute x = 1 into the equation to find f'(1). First, evaluate y when x = 1, which is f(1), to substitute into the equation for dy/dx. f(1) = (2*1 - 1)^15 * (5*1 - 4)^5 * (9*1 - 8)^5 / (11*1 - 10)^2 * (14*1 - 13)^3 * (17*1 - 16)^3 f(1) = (1)^15 * (1)^5 * (1)^5 / (1)^2 * (1)^3 * (1)^3 f(1) = 1 Now we substitute x = 1 and f(1) = 1 into the derivative equation: f'(1) = 1 * [ 30/(2*1 - 1) + 25/(5*1 - 4) + 45/(9*1 - 8) - 22/(11*1 - 10) - 42/(14*1 - 13) - 51/(17*1 - 16) ] f'(1) = [ 30/1 + 25/1 + 45/1 - 22/1 - 42/1 - 51/1 ] f'(1) = 30 + 25 + 45 - 22 - 42 - 51 f'(1)= 35 Therefore, using logarithmic differentiation, we find that f'(1) = 35.
Derivative Calculation of a Quadratic Function
The image shows a mathematical problem where you are given a function f(x) = 3x^2 + 2x and asked to find the derivative of the function at x = 2, represented as f'(2). To find the derivative f'(x) of the function f(x) = 3x^2 + 2x, we will use the power rule for differentiation. The power rule states that the derivative of x^n is n*x^(n-1). So for f(x) = 3x^2 + 2x: f'(x) = d/dx (3x^2) + d/dx (2x) = 3 * 2x^(2-1) + 2 * 1x^(1-1) = 6x + 2 Now we need to evaluate the derivative at x = 2: f'(2) = 6(2) + 2 = 12 + 2 = 14 Therefore, the derivative of the function at x = 2, f'(2), is 14.
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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