Question

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Differentiate the following with respect to x:
5x⁴+4x^3/4-3x²+2x​

Answer 1

$\displaystyle\underline{\text{Solution :}}$

$\displaystyle\mathrm{Now,\:\frac{d}{dx}\bigg[\frac{5x^{4}+4x^{3}}{4-3x^{2}+2x}\bigg]}$

$\displaystyle\mathrm{=\frac{(4-3x^{2}+2x)\frac{d}{dx}(5x^{4}+4x^{3})-(5x^{4}+4x^{3})\frac{d}{dx}(4-3x^{2}+2x)}{(4-3x^{2}+2x)^{2}}}$

$\displaystyle\mathrm{=\frac{(4-3x^{2}+2x)(20x^{3}+12x^{2})-(5x^{4}+4x^{3})(-6x+2)}{(4-3x^{2}+2x)^{2}}}$

$\displaystyle\mathrm{=\frac{(80x^{3}+48x^{2}-60x^{5}-36x^{4}+40x^{4}+24x^{3})-(-30x^{5}+10x^{4}-24x^{4}+8x^{3})}{(4-3x^{2}+2x)^{2}}}$

$\displaystyle\mathrm{=\frac{(104x^{3}+48x^{2}+4x^{4}-60x^{5})-(-30x^{5}-14x^{4}+8x^{3})}{(4-3x^{2}+2x)^{2}}}$

$\displaystyle\mathrm{=\frac{104x^{3}+48x^{2}+4x^{4}-60x^{5}+30x^{5}+14x^{4}-8x^{3}}{(4-3x^{2}+2x)^{2}}}$

$\displaystyle\mathrm{=\frac{96x^{3}+48x^{2}+18x^{4}-30x^{5}}{(4-3x^{2}+2x)^{2}}}$

$\displaystyle\mathrm{=\frac{6x^{2}(16x+8+3x^{2}-5x^{3})}{(4-3x^{2}+2x)^{2}}}$

$\displaystyle\to \boxed{\mathrm{\frac{d}{dx}\bigg[\frac{5x^{4}+4x^{3}}{4-3x^{2}+2x}\bigg]=\frac{6x^{2}(16x+8+3x^{2}-5x^{3})}{(4-3x^{2}+2x)^{2}}}}$

$\displaystyle\underline{\text{Rule :}}$

$\displaystyle\mathrm{\frac{d}{dx}\frac{u}{v}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}}$

$\displaystyle\text{where u, v are functions of x.}$