Question

(x2-3)в 5 степени найти производную

Answer 1

$((x^2-3)^5)'=5*(x^2-3)^{5-1}*(x^2-3)'=5(x^2-3)^4*((x^2)'-(3)')=$

$5(x^2-3)^4*(2x-0)=5(x^2-3)^4*2x=10x(x^2-3)^4$

Answer 2

To find the derivative of the function f(x) = (x^2 - 3)^5 with respect to x, we'll use the chain rule. The chain rule states that if you have a composite function y = g(h(x)), then the derivative of y with respect to x is given by dy/dx = g'(h(x)) * h'(x).

Let's apply the chain rule step by step:

Step 1: Identify the inner function and the outer function. Inner function: u = x^2 - 3 Outer function: y = u^5

Step 2: Find the derivatives of the inner and outer functions. dy/du = 5u^4 (derivative of u^5 with respect to u) du/dx = 2x (derivative of x^2 - 3 with respect to x)

Step 3: Apply the chain rule formula: dy/dx = dy/du * du/dx

dy/dx = 5u^4 * 2x

Step 4: Substitute back the value of u. u = x^2 - 3

dy/dx = 5(x^2 - 3)^4 * 2x

So, the derivative of f(x) = (x^2 - 3)^5 with respect to x is dy/dx = 5(x^2 - 3)^4 * 2x.