Differentiate the following functions:

Solution:

Formula:

$\frac{d}{d x} \frac{u}{v}=\frac{v \frac{d}{d x} u-u \frac{d}{d x} v}{u^{2}}$

(i) $\frac{3 x^{2}+4 x-5}{x}$

Applying, quotient rule

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

$=\frac{x(6 x+4)-\left(3 x^{2}+4 x-5\right) 1}{x^{2}}$

$=\frac{6 x^{2}+4 x-\left(3 x^{2}+4 x-5\right)}{x^{2}}$

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

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

Applying, quotient rule

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

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

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

$=\frac{x^{2}\left\{x^{3}+1+3 x^{3}-6 x^{2}\right\}-2\left(x^{4}+x\right)(x-2)}{x^{4}}$

$=\frac{4 x^{5}-6 x^{4}+x^{2}-2\left(x^{5}-2 x^{4}+x^{2}-2 x\right)}{x^{4}}$

$=\frac{2 x^{5}-2 x^{4}-x^{2}+4 x}{x^{4}}$

(iii) $\frac{x-4}{2 \sqrt{x}}$

Applying, quotient rule

$\frac{d}{d x} \frac{x-4}{2 \sqrt{x}}=\frac{2 \sqrt{x} \frac{d}{d x}(x-4)-(x-4) \frac{d}{d x} 2 \sqrt{x}}{4 x}$

$=\frac{2 \sqrt{x}-(x-4) 2 \frac{1}{2} x^{-\frac{1}{2}}}{4 x}$

$=\frac{2 \sqrt{x}-(x-4) x^{-\frac{1}{2}}}{4 x}$

$=\frac{2 \sqrt{x}-x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}}{4 x}$

$=\frac{\sqrt{x}+4 x^{-\frac{1}{2}}}{4 x}$

(iv) $\frac{(1+x) \sqrt{x}}{\sqrt[3]{x}}$

Applying, quotient rule

$\frac{d}{d x} \frac{(1+x) \sqrt{x}}{\sqrt[3]{x}}=\frac{\sqrt[3]{x} \frac{d}{d x}(1+x) \sqrt{x}-(1+x) \sqrt{x} \frac{d}{d x} \sqrt[3]{x}}{x^{\frac{2}{3}}}$

$=\frac{\sqrt[3]{x}\left\{(1+x) \frac{d}{d x} \sqrt{x}+\sqrt{x} \frac{d}{d x}(1+x)\right\}-(1+x) \sqrt{x} * \frac{1}{3} x^{\frac{-2}{3}}}{x^{\frac{2}{3}}}$

$=\frac{\sqrt[3]{x}\left\{(1+x) * \frac{1}{2} x^{-\frac{1}{2}}+\sqrt{x}\right\}-(1+x) * \frac{1}{3} x^{\frac{-1}{6}}}{x^{\frac{2}{3}}}$

$=\frac{\sqrt[3]{x}\left\{\frac{1}{2} x^{-\frac{1}{2}}+\frac{1}{2} x^{\frac{1}{2}}+\sqrt{x}\right\}-\frac{1}{3}\left(x^{\frac{-1}{6}}+x^{\frac{5}{6}}\right)}{x^{\frac{2}{3}}}$

$=\frac{\sqrt[3]{x}\left\{\frac{1}{2} x^{-\frac{1}{2}}+\frac{1}{2} x^{\frac{1}{2}}+\sqrt{x}\right\}-\frac{1}{3}\left(x^{\frac{-1}{6}}+x^{\frac{5}{6}}\right)}{x^{\frac{2}{3}}}$

$=\frac{\frac{1}{2} x^{-\frac{1}{6}}+\frac{1}{2} x^{\frac{5}{6}}+\sqrt{x}-\frac{1}{3}\left(x^{\frac{-1}{6}}+x^{\frac{5}{6}}\right)}{x^{\frac{2}{3}}}$

$=\frac{\frac{1}{6} x^{-\frac{1}{6}}+\frac{1}{6} x^{\frac{5}{6}}+\sqrt{x}}{x^{\frac{2}{3}}}$

(v) $\frac{a x^{2}+b x+c}{\sqrt{x}}$

Applying, quotient rule

$\frac{d}{d x} \frac{a x^{2}+b x+c}{\sqrt{x}}=\frac{\sqrt{x} \frac{d}{d x}\left(a x^{2}+b x+c\right)-\left(a x^{2}+b x+c\right) \frac{d}{d x} \sqrt{x}}{x}$

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

$=\frac{\frac{3}{2} a x^{\frac{3}{2}}+\frac{1}{2} b x^{\frac{1}{2}}-\frac{1}{2} c x^{-\frac{1}{2}}}{x}$

(vi) $\frac{a+b \cos x}{\sin x}$

Applying, quotient rule

$\frac{d}{d x} \frac{a+b \cos x}{\sin x}=\frac{\sin x \frac{d}{d x}(a+b \cos x)-(a+b \cos x) \frac{d}{d x} \sin x}{\sin ^{2} x}$

$=\frac{\sin x(-b \sin x)-(a+b \cos x) \cos x}{\sin ^{2} x}$

$=\frac{-b \sin ^{2} x-a \cos x-b \cos ^{2} x}{\sin ^{2} x}$

$=\frac{-b(1)-a \cos x}{\sin ^{2} x}$