How do you differentiate $ln ({sec}^{2} \cdot x)$ $ln(sec^2 * x)$ ?

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How do you differentiate $ln ({sec}^{2} \cdot x)$ $ln(sec^2 * x)$ ?

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Answer 1 · 2018-07-05T08:31:32

$\frac{d y}{d x} = 2 tan x$ $(dy)/(dx)=2tanx$

Explanation:

Here ,

$y = ln ({sec}^{2} x)$ $y=ln(sec^2x)$

Let ,

$y = ln u, w h e r e, u = {sec}^{2} x$ $y=lnu , where , u=sec^2x$

$\frac{d y}{d u} = \frac{1}{u} and \frac{d u}{d x} = 2 sec x \cdot sec x tan x = 2 {sec}^{2} x tan x$ $(dy)/(du)=1/u and (du)/(dx)=2secx*secxtanx=2sec^2xtanx$

Diff.w.r.t. $x$ $x$ using Chain Rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)$

$\frac{d y}{d x} = \frac{1}{u} \times 2 {sec}^{2} x tan x$ $(dy)/(dx)=1/u xx 2sec^2xtanx$

Subst, back , $u = {sec}^{2} x$ $u=sec^2x$

$:.(dy)/(dx)=1/sec^2x xx 2sec^2xtanx$

$(dy)/(dx)=2tanx$

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How do you differentiate $ln ({sec}^{2} \cdot x)$ $ln(sec^2 * x)$ ?

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