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Differentiation With Respect to a Function

April 10th, 2009 |

Author: Merrin

Leibniz’s notation is quite powerful. It allows us to imagine taking the derivative not with respect to x, but perhaps with respect to another function. Differentiation with respect to a function is an elegant application of differentiation.

Example 1. Find the derivative of x with respect to the square root of x.

$\cfrac{d}{d\sqrt{x} } \,x$

Solution 1. We can solve for the derivative by making the argument look like a function of the square root of x.

$\cfrac{d}{d\sqrt{x} }\, x=\cfrac{d}{d\sqrt{x} } \left(\sqrt{x} \right)^{2} =\cfrac{d}{du} u^{2} =2u=2\sqrt{x}$

Alternatively, we can use a chain rule on the differentiation operator to get derivatives with respect to x alone.

$d\sqrt{x} =\cfrac{1}{(2\sqrt{x} )} dx \\ \\\cfrac{d}{d\sqrt{x} } =\cfrac{1}{1/(2\sqrt{x} )} \cfrac{d}{dx} =2\sqrt{x} \cfrac{d}{dx} \\ \\ \cfrac{d}{d\sqrt{x} }\, x=2\sqrt{x} \cfrac{d}{dx} x=2\sqrt{x}$

Example 2. Differentiation with respect to other functions is also possible. Find the following derivative.

$\cfrac{d}{d(\sin x)} (\tan x)$

Solution 2. We use the first method from the previous example to take the derivative.

$\cfrac{d}{d(\sin x)} \tan x = \cfrac{d}{d(\sin x)} \cfrac{\sin x} {\sqrt{1-(\sin x)^{2} } } = \cfrac{d}{du}\cfrac{u} {\sqrt{1-u^{2}}}$
$\cfrac{1}{\sqrt{1-u^2}}+\cfrac{u^2}{(1-u^2)^{3/2}}=\cfrac{1}{(1-u^2)^{3/2}}=\sec^3x$

We can also try the second method from the previous example which uses the chain rule on the differentiation operator.

$\cfrac{d}{d(\sin x)} (\tan x)=\cfrac{1}{\cos x} \cfrac{d}{dx} \tan x=\sec ^{3} x$

Some freaky derivatives are possible when differentiating with respect to functions. There are some situations which actually make use of this sort of calculation, if not as a regular derivative, then even as a partial derivative.

$\cfrac{d}{d[u(x)]}f(x) = \cfrac{d}{du} f(u) = \cfrac{1}{u'(x)}\cfrac{d}{dx}f(x)$