4

I've been trying to find a way to take the derivative of one expression with respect to another expression, but I can't think of any obvious way to do it, and I'm not even sure where I should start.

For example, would it be possible to find the derivative of $x^6$ with respect to $x^2$, or the derivative of $\sin(x)$ with respect to $cos(x)$?

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
Anderson Green
  • 1,035
  • 2
    You could do something like: $f(x) = x^6 = (x^2)^3 \rightarrow f'(x) = 6x^5 = 3(x^2)^2(2x) = 6x^5$ (the latter is using the product and chain rule). – Amzoti Sep 12 '13 at 01:38

3 Answers3

10

If one knows the chain rule, $$\frac{df}{dg}\frac{dg}{dx}=\frac{df}{dx},$$ then one may solve for $$\frac{df}{dg}=\frac{\frac{df}{dx}}{\frac{dg}{dx}}.$$ For your one of your examples, $\frac{d\sin x}{d\cos x}$, we may compute this to be $$\frac{d\sin x}{d\cos x}=\frac{\frac{d \sin x}{dx}}{\frac{d\cos x}{dx}}=\frac{\cos x}{-\sin x}=-\cot x.$$

One should also note that we should define the derivative of $f$ with respect to $g$ as the following limit:

$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{g(x+h)-g(x)}.$$

Baby Dragon
  • 2,823
  • 22
  • 32
  • The example that I posted was "the derivative of sin(x) with respect to cos(x)", and not "the derivative of sin(x) divided by the derivative of cos(x)", so I'm not sure if this solution is correct. – Anderson Green Sep 12 '13 at 02:46
  • 1
    My solution is correct, it just is not the example you posted. At any rate I edited it so it is your example. – Baby Dragon Sep 12 '13 at 03:32
1

$$ \frac{d\ x^6}{d\ x^2} = \frac{\frac{d\ x^6}{d\ x}}{\frac{d\ x^2}{d\ x}} = \frac{6x^5}{2x} = 3x^4 $$

Ertxiem - reinstate Monica
  • 1,664
  • 3
  • 11
  • 15
Bilal Ahmed
  • 11
0

In addition to Baby Dragon's answer which works for the general case, here is another approach that works for the trig case.

Let $z = \cos x$.

$\frac{d}{d \cos x} \sin x=\frac{d}{dz}\sqrt{1-z^2}=\frac{-z}{\sqrt{1-z^2}}=\frac{-\cos x}{\sin x}.$

Student
  • 398
  • 1
  • 13