Question

write the derivative of sim x with respect to x

Answer 1

We use the chain rule:

[math]\displaystyle \frac{\mathrm d(\sin x)}{\mathrm d(\cos x)} = \frac{\mathrm d(\sin x)}{\mathrm dx} \cdot \frac{\mathrm dx} {\mathrm d(\cos x)}[/math]

We know the derivatives of [math]\sin[/math] and [math]\cos[/math]:

[math]\displaystyle \frac{\mathrm d(\sin x)}{\mathrm dx} = \cos x[/math]

and

[math]\displaystyle \frac{\mathrm d(\cos x)}{\mathrm dx} = -\sin x[/math]

Also by the chain rule, we have:

[math]\displaystyle \frac{\mathrm dx} {\mathrm d(\cos x)} = -\frac 1 {\sin x}[/math]

So we have:

[math]\displaystyle \frac{\mathrm d(\sin x)}{\mathrm d(\cos x)} = -\frac{\cos x} {\sin x} = \boxed{-\cot x}[/math]

You can probably derive this geometrically with the unit circle but I’ll just leave you with this.

Answer 2

Answer:

Step-by-step explanation:

hiii mate..

here is ur answer..

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