Differentiate the following functions:

Solution:

Formula:

$\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{dg}} \mathrm{f}(\mathrm{g}) \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{g}$

Chain rule -

$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{uv})=\mathrm{u} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{v}+\mathrm{v} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u}$

Where u and v are the functions of x.

(i) $(2 x+3)(3 x-5)$

Applying, Chain rule

Here, $u=2 x+3$

$V=3 x-5$

$\frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x}+3)(3 \mathrm{x}-5)=(2 \mathrm{x}+3) \frac{\mathrm{d}}{\mathrm{dx}}(3 \mathrm{x}-5)+(3 \mathrm{x}-5) \frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x}+3)$

$=(2 x+3)\left(3 x^{1-1}+0\right)+(3 x-5)\left(2 x^{1-1}+0\right)$

$=6 x+9+6 x-10$

$=12 x-1$

(ii) $x(1+x)^{3}$

Applying, Chain rule

Here, u = x

$V=(1+x)^{3}$

$\frac{d}{d x} \times(1+x)^{3}=x \frac{d}{d x}(1+x) 3+(1+x) 3 \frac{d}{d x}(x)$

$=x \times 3 \times(1+x)^{2}+(1+x)^{3}(1)$

$=(1+x)^{2}(3 x+x+1)$

$=(1+x)^{2}(4 x+1)$

(iii) $\left(\sqrt{x}+\frac{1}{x}\right)\left(x-\frac{1}{\sqrt{x}}\right)=\left(x^{1 / 2}+x^{-1}\right)\left(x-x^{-1 / 2}\right)$

Applying, Chain rule

Here, $u=\left(x^{1 / 2}+x^{-1}\right)$

$V=\left(X-X^{-1 / 2}\right)$

$\frac{d}{d x}\left(x^{1 / 2}+x^{-1}\right)\left(x-x^{-1 / 2}\right)$

$=\left(x^{1 / 2}+x^{-1}\right) \frac{d}{d x}\left(x-x^{-1 / 2}\right)+\left(x-x^{-1 / 2}\right) \frac{d}{d x}\left(x^{1 / 2}+x^{-1}\right)$

$=\left(x^{1 / 2}+x^{-1}\right)\left(1+\frac{1}{2} x^{-3 / 2}\right)+\left(x-x^{-1 / 2}\right)\left(\frac{1}{2} x^{-1 / 2}-x^{-2}\right)$

$=x^{1 / 2}+x^{-1}+\frac{1}{2} x^{-1}+\frac{1}{2} x^{-5 / 2}+\frac{1}{2} x^{1 / 2}-x^{-1}-\frac{1}{2} x^{-1}+x^{-5 / 2}$

$=\frac{3}{2} x^{1 / 2}+\frac{3}{2} x^{-5 / 2}$

(iv)

$\left(x-\frac{1}{x}\right)^{2}$

Differentiation of composite function can be done by

$\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{dg}} \mathrm{f}(\mathrm{g}) \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{g}$

Here, $f(g)=g^{2}, g(x)=x-\frac{1}{x}$

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{2}=2 \mathrm{~g} \times\left(1+\frac{1}{\mathrm{x}^{2}}\right)$

$=2\left(x-\frac{1}{x}\right)\left(1+\frac{1}{x^{2}}\right)$

$=2\left(x+\frac{1}{x}-\frac{1}{x}+\frac{1}{x^{3}}\right)$

$=2\left(x+\frac{1}{x^{3}}\right)$

(v)

$\left(x^{2}-\frac{1}{x^{2}}\right)^{3}$

Differentiation of composite function can be done by

$\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{dg}} \mathrm{f}(\mathrm{g}) \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{g}$

Here, $f(g)=g^{3}, g(x)=x^{2}-\frac{1}{x^{2}}$

$\frac{d}{d x}\left(x^{2}-\frac{1}{x^{2}}\right)^{3}=3 g^{2} \times\left(2 x-\frac{2}{x^{3}}\right)$

$=3\left(x^{2}-\frac{1}{x^{2}}\right)^{2}\left(2 x-\frac{2}{x^{3}}\right)$

$=3\left(2 x^{3}-\frac{2}{x}-\frac{2}{x}+\frac{2}{x^{5}}\right)$

$=3\left(2 x^{3}-\frac{4}{x}+\frac{2}{x^{5}}\right)$

(vi) $\left(2 x^{2}+5 x-1\right)(x-3)$

Applying, Chain rule

Here, $u=\left(2 x^{2}+5 x-1\right)$

$V=(x-3)$

$\frac{d}{d x}\left(2 x^{2}+5 x-1\right)(x-3)$

$=\left(2 x^{2}+5 x-1\right) \frac{d}{d x}(x-3)+(x-3) \frac{d}{d x}\left(2 x^{2}+5 x-1\right)$

$=\left(2 x^{2}+5 x-1\right) \times 1+(x-3)(4 x+5)$

$=2 x^{2}+5 x-1+4 x^{2}-7 x-15$

$=6 x^{2}-2 x-16$