Differentiate $sin x$ $sinx$ $/$ $/$ $5 x$ $5x$ + ${sec}^{2} x$ $sec^2 x"$ ?

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Differentiate $sin x$ $sinx$ $/$ $/$ $5 x$ $5x$ + ${sec}^{2} x$ $sec^2 x"$ ?

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Answer 1 · 2017-06-06T15:01:38

$\frac{x cos x - sin x}{5 x^{2}} + 2 {sec}^{2} x tan x$ $(xcosx-sinx)/(5x^2)+2sec^2xtanx$

Explanation:

$\frac{d}{d x} (\frac{sin x}{5 x} + {sec}^{2} x)$ $d/dx(sinx/(5x)+sec^2x)$
$= \frac{d}{d x} (\frac{sin x}{5 x}) + \frac{d}{d x} ({sec}^{2} x)$ $=d/dx(sinx/(5x))+d/dx(sec^2x)$
$= \frac{\frac{d (sin x)}{d x} (5 x) - sin x \frac{d (5 x)}{d x}}{{(5 x)}^{2}} + \frac{d ({sec}^{2} x)}{d sec x} \cdot \frac{d (sec x)}{d x}$ $=((d(sinx))/dx(5x)-sinx(d(5x))/dx)/(5x)^2+(d(sec^2x))/(dsecx)*(d(secx))/dx$
$= \frac{cos x \cdot 5 x - 5 sin x}{25 x^{2}} + 2 sec x \cdot sec x tan x$ $=(cosx*5x-5sinx)/(25x^2)+2secx*secxtanx$

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Differentiate $sin x$ $sinx$ $/$ $/$ $5 x$ $5x$ + ${sec}^{2} x$ $sec^2 x"$ ?

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