Ncert Solutions Maths class 12th

68. Let y = cos (log x + e^x)

$\therefore \frac{dy}{dx}=\frac{d}{dx}$cos (log x + e^x)

= - sin (log x + e^x) $\frac{d}{dx}$ (log x + e^x)

= - sin (log x + e^x)  $\left[\frac{1}{x}+e^x\right]$

$=-\left[\frac{1}{x}+e^x\right]$ sin (log x + e^x).

67. $y=\frac{cotx}{logx}$

$\therefore \frac{dy}{dx}=\frac{logx\frac{d}{da}cox-cosx\frac{d}{dx}logx}{{[logx]}^2}$

$=\frac{-sinx·logx-cosx*\frac{1}{x}}{{[logx]}^2.}$

$=\frac{-[xsinxlogx-cosx]}{x{(logx)}^2.}$

 66. Let y = log (log x)

$\therefore \frac{dy}{dx}=\frac{1}{logx}\frac{d}{dx}(logx)$

$=\frac{1}{logx}*\frac{1}{x}$

$\frac{dy}{dx}=\frac{1}{xlogx}$

65. Kindly go through the solution

64. Let y = e^x + e^x2 + … + e^x5.

$\therefore \frac{dy}{dx}=\frac{d}{dx}$ (e^x + e^x2 + e^x3 + e^x4 + e^x5).

$=\frac{d{e}^x}{dx}+\frac{d{e}^{x^2}}{dx}+\frac{d{e}^{x^3}}{dx}+\frac{d}{dx}{e}^{x^4}+\frac{d}{dx}{e}^{x^{5.}}$

$=e^x+e^{x^2}\frac{d{x}^2}{dx}+e^{x^3}\frac{d{x}^3}{dx}+e^{x^4}\frac{d{x}^4}{dx}+e^{x^5}\frac{d{x}^5}{dx}.$

= e^x + 2x e^x2 + 3x²e^x3 + 4x³e^x4 + 5ee^x4

63. Let y = log (cos e^x).

$\therefore \frac{dy}{dx}=\frac{d}{d\mathrm{x }}$log (cos e^x)

$=\frac{1}{cos{e}^x}\frac{d}{dx}$ (cos e^x)

= - $\frac{}{}$$\frac{sin{e}^x}{cos{e}^x}\frac{d{e}^{x.}}{dx}$

= -e^x [tan e^x]

62. Kindly go through the solution

61.  Kindly go through the solution

Let y = e^x3

60. Kindly go through the solution

59. Let y = $\frac{e^x}{sinx}$

$\therefore \frac{dy}{dx}=\frac{sinx\frac{d{e}^x}{dx}-e^x\frac{d}{dx}sinx}{si{n}^{2}x}$

$=\frac{sinx{e}^x-e^{x}cosx}{si{n}^{2}x}$

$=e^x\frac{(sinx-cosx)}{si{n}^{-2}x.}$