Differentiate following w.r.t x Exercise 1.1 XIII

 Differentiate the following w.r.t x Exercise 1.1

3)Differentiate the following w.r.t x :

(xiii)$log\left[\sqrt{\frac{1-cos3x}{1+cos3x}}\right]$

Solution-

$\frac{dy}{dx}=\frac{d}{dx}\left(log\left[\sqrt{\frac{1-cos3x}{1+cos3x}}\right]\right)$

$=\frac{1}{\sqrt{\frac{1-cos3x}{1+cos3x}}}\frac{d}{dx}\left(\sqrt{\frac{1-cos3x}{1+cos3x}}\right)$

$=\frac{1}{\sqrt{\frac{1-cos3x}{1+cos3x}}}\frac{1}{2\sqrt{\frac{1-cos3x}{1+cos3x}}}\frac{d}{dx}\left(\frac{1-cos3x}{1+cos3x}\right)$

$=\frac{1}{\sqrt{\frac{1-cos3x}{1+cos3x}}}\frac{1}{2\sqrt{\frac{1-cos3x}{1+cos3x}}}\frac{(1-cos3x)\frac{d}{dx}(1+cos3x)+(1+cos3x)\frac{d}{dx}(1-cos3x)}{(1+cos3x)^2}$

$=\frac{1}{2\left(\frac{1-cos3x}{1+cos3x}\right)}\frac{(1-cos3x)(0+sin3x\frac{d}{dx}3x)+(1+cos3x)(0+sin3x\frac{d}{dx}3x)}{(1+cos3x)^2}$

$=\frac{1+cos3x}{2(1-cos3x)}\frac{(1-cos3x)(3sin3x)+(1+cos3x)(3sin3x)}{(1+cos3x)^2}$

$=\frac{6sin3x(1+cos3x)}{2(1-cos3x)^3}$

(i)$(x^2+4x+1)^3+(x^3-5x-2)^4$

(ii)$(1+4x)^5(3x+x-x^2)^8$

(iii)$\frac{x}{\sqrt{7-3x}}$

(iv)$\frac{(x^3-5)^5}{(x^3+5)^3}$

(v)$(1+sin^2x)^2+(1+cos^2)^3$

(vi)$\sqrt{cos x}+\sqrt{cos\sqrt{x}}$

(vii)$log(sec 3x+tan 3x)$

(viii)$\frac{1+sin x^{\circ}}{1-sin x^{\circ}}$

(x)$\frac{e^{2x}-e^{-2x}}{e^{2x}+e{-2x}}$

(xi)$\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}$

(xii)$log\left[ tan^3xsin^4x(x^2+7)\right]$

(xiii)$log\left[\sqrt{\frac{1-cos3x}{1+cos3x}}\right]$

(xiv)$log\left[\sqrt{\frac{1+cos\frac{5x}{2}}{1-cos\frac{5x}{2}}}\right]$

(xv)$log\left[\sqrt{\frac{1-sinx}{1+sinx}}\right]$

Differentiate following w.r.t x Exercise 1.1 XIII