Differentiate the following w.r.t x Exercise 1.1
3)Differentiate the following w.r.t x :
(xii)$log\left[ tan^3xsin^4x(x^2+7)\right]$
Solution-
$\frac{dy}{dx}=\frac{d}{dx}\left(log\left[ tan^3xsin^4x(x^2+7)\right]\right)$
$=\frac{1}{ tan^3xsin^4x(x^2+7)}\frac{d}{dx}tan^3xsin^4x(x^2+7)$
$=\frac{1}{ tan^3xsin^4x(x^2+7)}\left(tan^3xsin^4x\frac{d}{dx}(x^2+7)+tan^3x(x^2+7)\frac{d}{dx}sin^4x+sin^4x(x^2+7)\frac{d}{dx}tan^3x\right)$
$=\frac{1}{ tan^3xsin^4x(x^2+7)}\left(tan^3xsin^4x(2x)+tan^3x(x^2+7)4sin^3x\frac{d}{dx}(sinx)+sin^4x(x^2+7)3tan^2x\frac{d}{dx}(tanx)\right)$
$=\frac{1}{tan^3xsin^4x(x^2+7)}\left(2xtan^3xsin^4x+tan^3x(x^2+7)4sin^3xcosx+3sin^4x(x^2+7)tan^3xsecx\right)$
(iv)$\frac{(x^3-5)^5}{(x^3+5)^3}$
(xiv)$log\left[\sqrt{\frac{1+cos\frac{5x}{2}}{1-cos\frac{5x}{2}}}\right]$
(xv)$log\left[\sqrt{\frac{1-sinx}{1+sinx}}\right]$
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| Differentiate following w.r.t x Exercise 1.1 XIl |




