Differentiation (Part-38) | S N De

 

$~10(i)~~(2x^2+3)^4$

Sol. let $~~y=(2x^2+3)^4 \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[(2x^2+3)^4]\\=4(2x^2+3)^3\cdot \frac{d}{dx}(2x^2+3)\\=4(2x^2+3)^3\cdot[\frac{d}{dx}(2x^2)+\frac{d}{dx}(3)]\\=4(2x^2+3)^3\cdot[4x+0]\\=16x(2x^2+3)^3~~\text{(ans.)}$

$~(ii)~~\sqrt{x^2+a^2}$

Sol. let $~~~~~~y=\sqrt{x^2+a^2}=(x^2+a^2)^{1/2} \\ \therefore ~\frac{dy}{dx}\\=\frac{d}{dx}[(x^2+a^2)^{1/2}]\\=\frac 12(x^2+a^2)^{\frac 12-1}~\cdot \frac{d}{dx}(x^2+a^2)\\=\frac 12(x^2+a^2)^{-\frac 12}\cdot [\frac{d}{dx}(x^2)+\frac{d}{dx}(a^2)]\\=\frac 12(x^2+a^2)^{-\frac 12}\cdot[2x+0]\\=\frac{x}{\sqrt{x^2+a^2}}~~\text{(ans.)}$

$~(iii)~~\frac{1}{px+q}$

Sol. let $~~y=\frac{1}{px+q}=(px+q)^{-1} \\ \therefore~~\frac{dy}{dx}\\=\frac{d}{dx}\left[(px+q)^{-1}\right]\\=-1\cdot(px+q)^{-1-1}\cdot \frac{d}{dx}(px+q)\\=-(px+q)^{-2}\cdot \left[\frac{d}{dx}(px)+\frac{d}{dx}(q)\right]\\=-(px+q)^{-2}\cdot [p\cdot 1+0]\\=-\frac{p}{(px+q)^{2}}~~\text{(ans.)}$

$~(iv)~~(\sqrt{\log x})^3$

Sol. let $~~~~~~y=(\sqrt{\log x})^3 \\ \text{or,}~~y=(\log x)^{\frac 32}\\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}\left[(\log x)^{\frac 32}\right]\\=\frac 32(\log x)^{\frac 32-1}\cdot \frac{d}{dx}(\log x)\\=\frac 32(\log x)^{\frac 12} \cdot \frac 1x\\=\frac{3\sqrt{\log x}}{2x}~~\text{(ans.)}$

$~(v)~~(e^{2x})^4$

Sol. let $~~y=(e^{2x})^4 \\ \therefore~~\frac{dy}{dx}\\=4(e^{2x})^{4-1}\cdot \frac{d}{dx}(e^{2x})\\=4(e^{2x})^3 \cdot e^{2x}\cdot \frac{d}{dx}(2x)\\=4(e^{2x})^3\cdot (e^{2x}) \cdot 2\\=8(e^{2x})^4\\=8e^{8x}~~\text{(ans.)}$

$~(vi)~~10^{10x}$

Sol. let $~~y=10^{10x} \\ \therefore~~\frac{dy}{dx}\\=\frac{d}{dx}\left(10^{10x}\right)\\=10^{10x} \cdot \log_e10\cdot\frac{d}{dx}(10x)\\=10^{10x} \cdot (\log_e10)\cdot 10\\=10^{10x+1} \log_e10~~\text{(ans.)}$

$~(vii)~~e^{x^9}$

Sol. let $~~y=e^{x^9} \\ \therefore~~\frac{dy}{dx}\\=\frac{d}{dx}(e^{x^9})\\=e^{x^9} \cdot \frac{d}{dx}(x^9)\\=e^{x^9} \cdot 9x^8\\=9x^8e^{x^9}~~\text{(ans.)}$

$~(viii)~~2^{2x^2+5x}$

Sol. let $~~y=2^{2x^2+5x} \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}(2^{2x^2+5x})\\=2^{2x^2+5x} \cdot \log_e2 \cdot \frac{d}{dx}(2x^2+5x)\\=2^{2x^2+5x} \cdot \log_e2 \cdot [2~\frac{d}{dx}(x^2)+5~\frac{d}{dx}(x)]\\=2^{2x^2+5x} \cdot \log_e2 \cdot[2 \cdot 2x+5]\\=(4x+5)~\cdot2^{2x^2+5x}~\log_e2~~\text{(ans.)}$

$~(ix)~~\log(\cot x)$

Sol. let $~~y=\log(\cot x) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\log(\cot x)]\\=\frac{1}{\cot x} \cdot \frac{d}{dx}(\cot x)\\=\frac{1}{\cot x}\cdot (-\csc^2x)\\=-\frac{\sin x}{\cos x} \cdot \csc x \cdot \csc x\\=-\frac{(\sin x \cdot \csc x)}{\cos x} \cdot \csc x\\=-\frac{1}{\cos x} \cdot \csc x\\=-\sec x \cdot \csc x~~\text{(ans.)}$

$~(x)~~\log(x^2+a^2)$

Sol. let $~~y=\log(x^2+a^2) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\log(x^2+a^2)]\\=\frac{1}{x^2+a^2} \cdot \frac{d}{dx}(x^2+a^2)\\=\frac{1}{x^2+a^2} \cdot [\frac{d}{dx}(x^2)+\frac{d}{dx}(a^2)]\\=\frac{1}{x^2+a^2} \cdot (2x+0)\\=\frac{2x}{x^2+a^2}~~\text{(ans.)}$

$~(xi)~~\log(\log x)~~(x>1)$

Sol. let $~~y=\log(\log x)~ \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\log(\log x)~]\\=\frac{1}{\log x} \cdot \frac{d}{dx}(\log x)\\=\frac{1}{\log x} \cdot \frac 1x\\=\frac{1}{x \log x}~~\text{(ans.)}$

$~(xii)~~\log[\log(\log x)]$

Sol. let $~~y=\log[\log(\log x)] \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}~[\log[\log(\log x)]]\\=\frac{1}{\log(\log x)} \cdot \frac{d}{dx}[\log(\log x)]\\=\frac{1}{\log(\log x)} \cdot \frac{1}{\log x}\cdot \frac{d}{dx}(\log x)\\=\frac{1}{\log(\log x)} \cdot \frac{1}{\log x}\cdot\frac 1x\\=\frac{1}{x \log x \cdot \log(\log x)}~~\text{(ans.)}$

$~(xiii)~~10^{\log(\cos x)}$

Sol. let $~~y=10^{\log(\cos x)} \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}~[10^{\log(\cos x)}]\\=10^{\log(\cos x)} \cdot \log_e10 \cdot \frac{d}{dx}[\log(\cos x)]\\=10^{\log(\cos x)} \cdot \log_e10 \cdot \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)\\=10^{\log(\cos x)}\cdot \log_e10 \cdot \frac{1}{\cos x}\cdot (-\sin x)\\=-\tan x \cdot 10^{\log(\cos x)} \cdot \log_e10~~\text{(ans.)}$

$~(xiv)~~e^{e^x}$

Sol. let $~~y=e^{e^x} \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}(e^{e^x})\\=e^{e^x} \cdot \frac{d}{dx}(e^x)\\=e^{e^x} \cdot e^x\\=e^x\cdot e^{e^x}~~\text{(ans.)}$

$~(xv)~~\sin(\cos x^2)~~~~~\text{[NCERT]}$

Sol. let $~~y=\sin(\cos x^2)~\\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[\sin(\cos x^2)~]\\=\cos(\cos x^2)\cdot \frac{d}{dx}(\cos x^2)\\=\cos(\cos x^2) \cdot (-\sin x^2)\cdot \frac{d}{dx}(x^2)\\=\cos(\cos x^2) \cdot (-\sin x^2) \cdot (2x)\\=-2x\sin x^2\cdot \cos(\cos x^2)~~\text{(ans.)}$

$~(xvi)~~\sin x^{\circ}$

Sol. let $~~y=\sin x^{\circ}=\sin \frac{\pi x}{180}~~[*]\\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}\left[\sin \frac{\pi x}{180}\right]\\=\cos\left(\frac{\pi}{180}\right) \cdot \frac{d}{dx}\left(\frac{\pi x}{180}\right)\\=\cos\left(\frac{\pi}{180}\right) \cdot \frac{\pi}{180} \cdot \frac{d}{dx}(x)\\=\frac{\pi}{180} \cdot\cos x^{\circ}~~\text{(ans.)}$

Note $[*] :$ We know, $ ~~~180^{\circ}=\pi^{c} \\ \therefore 1^{\circ}=\frac{\pi^{c}}{180} \\ \therefore~x^{\circ}=\frac{\pi x}{180}$