How do you differentiate $y = 5 x^{6} - {sec}^{- 1} x$ $y=5x^6-sec^-1x$ ?

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How do you differentiate $y = 5 x^{6} - {sec}^{- 1} x$ $y=5x^6-sec^-1x$ ?

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Answer 1 · 2018-02-13T08:02:34

$\frac{d y}{d x} = 30 x^{5} - \frac{1}{x \sqrt{x^{2} - 1}}$ $dy/dx=30x^5-1/(xsqrt(x^2-1)$

Explanation:

$y = 5 x^{6} - {sec}^{- 1} x$ $y=5x^6-sec^-1x$
$\frac{d y}{d x} = \frac{d}{d x} (5 x^{6} - {sec}^{- 1} x)$ $dy/dx=d/dx(5x^6-sec^-1x)$
Using sum rule
$= \frac{d}{d x} (5 x^{6}) - \frac{d}{d x} ({sec}^{- 1} x)$ $=d/dx(5x^6)-d/dx(sec^-1x)$
$\frac{d}{d x} (5 x^{6}) = 5 . \frac{d}{d x} (x^{6})$ $d/dx(5x^6)=5.d/dx(x^6)$
$= (5) (6) x^{6 - 1} = 30 x^{5}$ $=(5)(6)x^(6-1)=30x^5$
$\frac{d}{d x} ({sec}^{- 1} x) = \frac{1}{x \sqrt{x^{2} - 1}}$ $d/dx(sec^-1x)=1/(xsqrt(x^2-1)$
Thus,
$\frac{d y}{d x} = 30 x^{5} - \frac{1}{x \sqrt{x^{2} - 1}}$ $dy/dx=30x^5-1/(xsqrt(x^2-1)$

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How do you differentiate $y = 5 x^{6} - {sec}^{- 1} x$ $y=5x^6-sec^-1x$ ?

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