Question

Find the derivatives of (a) cos²x and (b) sin³x but without using product or chain rule

Answer 1

The derivative of cos²x is -2sin(x)cos(x). The derivative of sin³x is 2sin(x)cos(x).

Step-by-step explanation:

(a) The derivative of cos²x can be found by applying the chain rule. Let u = cos(x), then cos²x = u².

Now, differentiate u² with respect to u, which is simply 2u. Finally, multiply by du/dx, which is -sin(x):

Derivative of cos²x = 2cos(x)(-sin(x)) = -2sin(x)cos(x)

(b) The derivative of sin³x can be found by using the trigonometric identity sin³x = (sinx)(sin²x). Now, differentiate sinx with respect to x, which is simply cos(x).

Then, differentiate sin²x with respect to x using the chain rule. Let u = sin(x), then sin²x = u².

The derivative of u² with respect to u is 2u, and multiply by du/dx, which is cos(x):

Derivative of sin³x = cos(x)(2sin(x)) = 2sin(x)cos(x)