Example Question - function differentiation
Here are examples of questions we've helped users solve.
Derivative of Exponential Function at Zero
题目要求求函数 g(t) = e^(t^2) 在 t = 0 处的第二十一次导数 g^(21)(0)。 我们知道指数函数 e^x 的Maclaurin级数展开式是 x 的全幂级数： e^x = 1 + x + x^2/2! + x^3/3! + ... + x^n/n! + ... 由此，我们可以推导出 g(t) = e^(t^2) 的 Maclaurin级数： g(t) = e^(t^2) = 1 + t^2 + t^4/2! + t^6/3! + ... + t^(2n)/n! + ... 接下来，为了找到 g(t) 在 t = 0 处的第二十一次导数，我们要找到级数中 t^(21) 对应的项，并计算系数。 在上述级数中，如果我们要找第二十一次导数，我们需要找到 t^21 这一项。然而在级数中只有偶数次幂，这意味着实际上并不存在 t^21 这一项，因此 g(t) 在 t = 0 处的第二十一次导数为 0。 所以，答案是 (B) 0。
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 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.
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