Question

Write the derivative of sinx with respect to cos x

Answer 1

d(sinx)d(cosx)=d(sinx)dx⋅dxd(cosx)

We know the derivatives of sin and cos:

d(sinx)dx=cosx

and

d(cosx)dx=−sinx

Also by the chain rule, we have:

dxd(cosx)=−1sinx

So we have:

d(sinx)d(cosx)=−cosxsinx=−cotx  

Answer 2

Answer:

The derivative of sin x with respect to cos x is - cot x.

Step-by-step explanation:

• Derivative of one function with respect to another function :
• Let, y = f(x) and z = g(x), then differentiation of y with respect to z is

$\frac{dy}{dz} = \frac{ \frac{dy}{dx} }{ \frac{dz}{dx} } = \frac{f'(x)}{g'(x)}$

• Let, y = sin x and z = cos x. Then,
• Derivative of sin x with respect to x is

$\frac{dy}{dx} = \frac{d \sin(x) }{dx} = \cos(x)$

• Derivative of cos x with respect to x is

$\frac{dz}{dx} = \frac{d \cos(x) }{dx} = - \sin(x)$

• Derivative of sin x with respect to cos x is

$\frac{d \sin(x) }{d \cos(x) } = \frac{ d\sin(x) }{dx} \times \frac{dx}{d \cos(x) } \\ = \frac{ \cos(x) }{ - \sin(x) } \\ = - \cot(x)$

• Hence, derivative of sin x with respect to cos x is - cot x .