Does it make sense to take the derivative of f(x) with respect to g(x)?

Question

Let $f(x) = \sin(x)$ and $g(x) = \cos(x)$.

Then I want to find the derivative of sin(x) with respect to cos(x).

So I have:

$$\frac{d\sin(x)}{d\cos(x)}$$

$d/dx \sin(x) = \cos(x)$, so $d\sin(x) = \cos(x)dx$.

$d/dx \cos(x) = -\sin(x)$, so $d\cos(x) = -\sin(x)dx$.

Then,

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

Is this legit or am I doing something wrong here?

Does this answer your question? Derivative of a function with respect to another function. — Deepak, Jan 26 '21 at 23:30

score 1 · Answer 1 · answered Jan 26 '21 at 23:32

1

We have:$$\frac{d(\sqrt{1-\cos^2x})}{d(\cos x)}$$Substitute $u=\cos(x)$:

$$\frac{d}{du}(\sqrt{1-u^2})$$ Can you continue from here?

answered Jan 26 '21 at 23:32

User

8,033

$\frac {-2u} {2\sqrt{1-u^2}}=\frac {-\cos x}{\sqrt{1-\cos^2 x}}=-\cot x \csc x$ – Vons Jan 27 '21 at 00:23
1

@Stacker it is $\frac{-\cos x}{\sqrt{1-\cos^2x}}=-\cot x$ – User Jan 27 '21 at 07:10
Ah. Thanks for pointing it out – Vons Jan 27 '21 at 09:31

score 0 · Answer 2 · answered May 07 '24 at 02:56

since none of the answers or comments pointed it out, I would like to add, that you can differentiate both the functions with respect to the same variable (x in this case, and then divide the equations)

ex:

$f(x) = sinx$ and $g(x) = cosx$

$\frac{d(f(x))}{dx} = cosx$ .....(1)

$\frac{d(g(x))}{dx} = -sinx$ .......(2)

dividing equation 1 by 2: $\frac{(1)}{(2)}$

$$\frac{\frac{d(f(x))}{dx}}{\frac{d(g(x))}{dx}} = \frac{cosx}{-sinx}$$

$$\frac{d(f(x))}{d(g(x))} = -cotx$$

Same method can also be used for functions more complex than this one

Does it make sense to take the derivative of f(x) with respect to g(x)?

2 Answers2