7.0 Differentiation of one function w.r.t other

$$ y = f\left( x \right)\;and\;z = g\left( x \right)\;then $$$$ \frac{{dy}}{{dx}} = \frac{d}{{dx}}f\left( x \right):derivative\;of\;y\;w.r.t\;x $$$$ \frac{{dz}}{{dx}} = \frac{d}{{dx}}g\left( x \right):derivative\;of\;z\;w.r.t\;x $$$$ where\;\frac{{dz}}{{dx}} \ne 0 $$ Thus,

$$\;\frac{{dy}}{{dz}} = \frac{{dy}}{{dx}} \bullet \frac{{dx}}{{dz}} = \frac{{dy}}{{dx}} \bullet \frac{1}{{\frac{{dz}}{{dx}}}} $$$$ \frac{{dy}}{{dz}} = \frac{{\frac{{dy}}{{dx}}}}{{\frac{{dz}}{{dx}}}} = \frac{{\frac{d}{{dx}}f\left( x \right)}}{{\frac{d}{{dx}}g\left( x \right)}} $$

Question 1: Find derivative of $x^5$ w.r.t $x^2$

Solution:

$$ Let\;y = {x^5}\;and\;z = {x^2}\;then $$$$ \frac{{dy}}{{dx}} = \frac{d}{{dx}}{x^5} $$$$\frac{{dz}}{{dx}} = \frac{d}{{dx}}{x^2}$$

We know that, $$\frac{d}{dx}x^n = nx^{n-1}$$$$\frac{d}{{dx}}{x^5} = 5{x^{5 - 1}} = 5{x^4}$$$$\frac{d}{{dx}}{x^2} = 2{x^{2 - 1}} = 2x$$$$ \frac{{dy}}{{dz}} = \frac{{\frac{{dy}}{{dx}}}}{{\frac{{dz}}{{dx}}}} = \frac{{\frac{d}{{dx}}{x^5}}}{{\frac{d}{{dx}}{x^2}}} = \frac{{5{x^4}}}{{2x}} = \frac{5}{2}{x^3} $$

Question 2: Find derivative of ${\tan ^{ - 1}}x\;w.r.t\;x$

Solution:

$$ Let\;y = {\tan ^{ - 1}}x\;and\;z = {x^2}\;then $$$$ \frac{{dy}}{{dx}} = \frac{d}{{dx}}{\tan ^{ - 1}}x $$$$ \frac{{dz}}{{dx}} = \frac{d}{{dx}}{x^2} $$

We know that, $$\frac{d}{dx} tan^{-1}x = \frac{1}{1+x^2}$$$$\frac{d}{dx}x^n = nx^{n-1}$$$$ \frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}} $$$$ \frac{d}{{dx}}{x^2} = 2{x^{2 - 1}} = 2x $$

Thus,

$$ \frac{{dy}}{{dz}} = \frac{{\frac{{dy}}{{dx}}}}{{\frac{{dz}}{{dx}}}} = \frac{{\frac{d}{{dx}}{{\tan }^{ - 1}}x}}{{\frac{d}{{dx}}{x^2}}} = \frac{{\frac{1}{{1 + {x^2}}}}}{{2x}} = \frac{1}{{2x\left( {1 + {x^2}} \right)}} $$