Differentiation (Part-38) | S N De

Find the derivatives w.r.t. $~x~~~ [1-4]$: 

$~1(i)~~ \log\left[e^x\left(\frac{x-2}{x+2}\right)^{3/4}\right]$

Sol. let $~~y=\log_e\left[e^x\left(\frac{x-2}{x+2}\right)^{3/4}\right]\\~~~~= \log_e e^x+\log_e \left(\frac{x-2}{x+2}\right)^{3/4} \\~~~~=x \log_e e+ \frac 34 \log_e\left(\frac{x-2}{x+2}\right)\\~~~~=x+ \frac 34[\log_e(x-2)-\log_e(x+2)] \\ \therefore \frac{dy}{dx}=\frac{d}{dx} (x)+ \frac 34 [\frac{d}{dx}~\log_e(x-2)\\~~~~~~~~-\frac{d}{dx}~\log_e(x+2)] \\~~~~~~~~~=1+\frac 34\left[\frac{1}{x-2}-\frac{1}{x+2}\right]\\~~~~~~~~~=1+\frac 34  \times  \frac{x+2-(x-2)}{(x-2)(x+2)}\\~~~~~~~~~=1+\frac 34 \times \frac{4}{x^2-2^2} \\~~~~~~~~~=1+\frac{3}{x^2-4}\\~~~~~~~~~= \frac{x^2-4+3}{x^2-4}\\~~~~~~~~~=\frac{x^2-1}{x^2-4}~~\text{(ans.)}$

$~(ii)~~\frac{5x}{\sqrt[3]{1-x^2}}+\sin^2(2x+3)$

Sol.  let $~~y=\frac{5x}{\sqrt[3]{1-x^2}}+\sin^2(2x+3)\\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}  \left(\frac{5x}{\sqrt[3]{1-x^2}}\right)+2\sin(2x+3) \\~~~\times  \frac{d}{dx}\{\sin(2x+3)\} \cdot \frac{d}{dx}(2x+3)\\=l+2\sin(2x+3)\cos(2x+3) \cdot 2 \\=l+2\sin [2(2x+3)]\\=l+2\sin(4x+6)\rightarrow(1)$

Now, $~~~~~l=\frac{d}{dx}\left(\frac{5x}{\sqrt[3]{1-x^2}}\right)\\~~~~~~~=5 \cdot\frac{d}{dx} \left[x(1-x^2)^{-1/3}\right]\\ \therefore \frac l5=1\cdot (1-x^2)^{-1/3}\\~~~+x \cdot \left(-\frac 13\right)(1-x^2)^{-1/3-1}\cdot (-2x)\\~~~~~~~=(1-x^2)^{-1/3}+\frac{2x^2}{3}(1-x^2)^{-4/3}\\~~~~~~~=(1-x^2)^{-1/3} \left(1+\frac{2x^2}{3} \cdot \frac{1}{1-x^2}\right)\\~~~~~~~=(1-x^2)^{-1/3} \cdot \frac{3(1-x^2)+2x^2}{3(1-x^2)}\\~~~~~~~=(1-x^2)^{-1/3} \cdot \frac{3-x^2}{3(1-x^2)} \\ \therefore l=\frac{5(3-x^2)}{3(1-x^2)^{4/3}}\rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ we get the required derivative w.r.t. $\,x\,$ is :

$~~\frac{5(3-x^2)}{3(1-x^2)^{4/3}}+2\sin(4x+6)$

$~(iii)~~\sqrt{1+\sin x}+\sqrt{\frac{1-x^2}{1+x^2}}$

Sol. let $~~y=\sqrt{1+\sin x}+\sqrt{\frac{1-x^2}{1+x^2}}\\=u+v~~\text{(say)} \\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} \rightarrow(1)$

Now, $~~~~1+\sin x\\=\cos^2(x/2)+\sin^2(x/2)\\~~~~+2\sin(x/2)\cos(x/2)\\=\left(\cos \frac x2+\sin\frac x2\right)^2 \\ \therefore \sqrt{1+\sin x}=\cos \frac x2+\sin \frac x2  \\ \text{so,}~~\frac{du}{dx}\\=\left(-\sin \frac x2 \right)\times  \frac 12+\left(\cos \frac x2 \right)\times \frac 12\\=\frac 12\left(\cos \frac x2-\sin \frac x2\right)\rightarrow(2)$

Also, $~v=\left(\frac{1-x^2}{1+x^2}\right)^{1/2} \\ \therefore \log v= \log\left(\frac{1-x^2}{1+x^2}\right)^{1/2} \\ \text{or,}~~ \log v= \frac 12\log \left(\frac{1-x^2}{1+x^2}\right)\\ \therefore \frac 1v~\frac{dv}{dx}\\=\frac 12 \cdot \frac{1}{\left(\frac{1-x^2}{1+x^2}\right)} \cdot \frac{d}{dx} \left(\frac{1-x^2}{1+x^2}\right)\\=\frac 12 \cdot \frac{1+x^2}{1-x^2}\cdot \frac{(1+x^2)(-2x)-(1-x^2)(2x)}{(1+x^2)^2}\\=\frac 12 \cdot \frac{1}{(1-x^2)(1+x^2)} \cdot (-2x)(1+x^2+1-x^2)\\=\frac{-2x}{(1+x^2)(1-x^2)} \\ \therefore \frac{dv}{dx}\\=\frac{-2x}{(1+x^2)(1-x^2)} \cdot \left(\frac{1-x^2}{1+x^2}\right)^{1/2}\\=\frac{-2x}{(1+x^2)} \cdot \frac{1}{\sqrt{1+x^2} \cdot \sqrt{1-x^2}}\\=\frac{-2x}{(1+x^2) \cdot \sqrt{(1+x^2)(1-x^2)}}\\=\frac{-2x}{(1+x^2)\sqrt{1-x^4}}\rightarrow(3)$

Hence, from $\,(1),\,(2),\,(3)~$ we get the required derivative w.r.t. $\,x\,$ is :

$~~~\frac 12\left(\cos \frac x2-\sin \frac x2\right)-\frac{2x}{(1+x^2)\sqrt{1-x^4}}$ 

$~(iv)~~~\frac{1}{\sqrt{x+a}-\sqrt{x+b}}+x^2\sin(x^3)$

Sol. let $~~y=\frac{1}{\sqrt{x+a}-\sqrt{x+b}}+x^2\sin(x^3)\\~~~=\frac{(\sqrt{x+a}+\sqrt{x+b})}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})}+x^2\sin(x^3)\\~~~=\frac{(\sqrt{x+a}+\sqrt{x+b})}{(\sqrt{x+a})^2-(\sqrt{x+b})^2}+x^2\sin(x^3)\\~~~=\frac{(\sqrt{x+a}+\sqrt{x+b})}{x+a-x-b}+x^2\sin(x^3)\\~~~=\frac{1}{(a-b)}~(\sqrt{x+a}+\sqrt{x+b})+x^2\sin(x^3)\\ \therefore~~\frac{dy}{dx}\\=\frac{1}{(a-b)}\left[\frac{d}{dx}(\sqrt{x+a})+\frac{d}{dx}(\sqrt{x+b})\right]\\~~~~~~~~~+x^2~\frac{d}{dx}(\sin x^3)+\sin (x^3) \frac{d}{dx}(x^2)\\=\frac{1}{(a-b)} \left(\frac{1}{2\sqrt{x+a}}+\frac{1}{2\sqrt{x+b}}\right)\\~~~~~~~~+x^2\cos(x^3)\cdot 3x^2+2x\sin(x^3)\\=\frac{1}{2(a-b)} \left(\frac{1}{\sqrt{x+a}}+\frac{1}{\sqrt{x+b}}\right)\\~~~~~~~~+3x^4\cos(x^3)+2x\sin(x^3)~~\text{(ans.)}$

$~(v)~~\log \sqrt{\frac{1+\cos^2x}{1-e^{}}}$

Sol. let $~~~y=\log \sqrt{\frac{1+\cos^2x}{1-e^{2x}}}=\log\left(\frac{1+\cos^2x}{1-e^{2x}}\right)^{1/2} \\ \therefore y=\frac 12 \log\left(\frac{1+\cos^2x}{1-e^{2x}}\right) \\ \text{or,}~~2y=\log\left(1+\cos^2x\right)-\log(1-e^{2x}) \\ \therefore~2~\frac{dy}{dx}\\=\frac{1}{1+\cos^2x}\cdot \frac{d}{dx}(1+\cos^2x)\\~~~~-\frac{1}{1-e^{2x}}\cdot \frac{d}{dx}(1-e^{2x})\\=\frac{1}{1+\cos^2x} \cdot 2\cos x(-\sin x)\\~~~~-\frac{1}{1-e^{2x}}\cdot (-2e^{2x})\\=\frac{-\sin2x}{1+\cos^2x}+\frac{2e^{2x}}{1-e^{2x}} \\ \therefore ~\frac{dy}{dx}=\frac{e^{2x}}{1-e^{2x}}-\frac{\sin2x}{2(1+\cos^2x)}~~\text{(ans.)}$

$~(vi)~~\tan^{-1}\left(\frac{\sin x}{1+\cos x}\right)$

Sol. let $~~y=\tan^{-1}\left(\frac{\sin x}{1+\cos x}\right)\\~~~~=\tan^{-1} \left(\frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}\right)\\~~~~=\tan^{-1}\left(\frac{\sin(x/2)}{\cos(x/2)}\right)\\~~~~=\tan^{-1}(\tan(x/2))\\~~~~=\frac x2 \\ \therefore \frac{dy}{dx}=\frac 12 \cdot \frac{d}{dx}(x)=\frac 12~\text{(ans.)}$