How do you find the derivative of $ln (({tan}^{2}) x)$ $ln((tan^2)x)$ ?

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How do you find the derivative of $ln (({tan}^{2}) x)$ $ln((tan^2)x)$ ?

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Answer 1 · 2017-08-16T17:19:30

$2 cot x + 2 tan x$ $2cotx+2tanx$ or $2 sec x csc x$ $2secxcscx$ , depending on preferred simplification

Explanation:

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Method 1 - No simplification

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

Note that $\frac{d}{d x} ln (x) = \frac{1}{x}$ $d/dxln(x)=1/x$ , so by the chain rule $\frac{d}{d x} ln (u) = \frac{1}{u} \cdot \frac{d u}{d x}$ $d/dxln(u)=1/u*(du)/dx$ . Then:

$\frac{d y}{d x} = \frac{1}{{tan}^{2} x} \cdot \frac{d}{d x} {tan}^{2} x$ $dy/dx=1/tan^2x*d/dxtan^2x$

And we need to use the chain rule again since we have a function squared:

$\frac{d y}{d x} = \frac{1}{{tan}^{2} x} \cdot 2 tan x \cdot \frac{d}{d x} tan x$ $dy/dx=1/tan^2x * 2tanx * d/dxtanx$

$\frac{d y}{d x} = \frac{1}{{tan}^{2} x} \cdot 2 tan x \cdot {sec}^{2} x$ $dy/dx=1/tan^2x * 2tanx * sec^2x$

$\frac{d y}{d x} = \frac{{cos}^{2} x}{{sin}^{2} x} \cdot \frac{2 sin x}{cos x} \cdot \frac{1}{{cos}^{2} x}$ $dy/dx=cos^2x/sin^2x * (2sinx)/cosx * 1/cos^2x$

$\frac{d y}{d x} = \frac{2}{sin x cos x}$ $dy/dx=2/(sinxcosx)$

$\frac{d y}{d x} = 2 sec x csc x$ $dy/dx=2secxcscx$

Method 2 - Simplification

Use the logarithm rules:

$ln (a^{b}) = b ln (a)$ $ln(a^b)=bln(a)$
$ln (a / b) = ln (a) - ln (b)$ $ln(a//b)=ln(a)-ln(b)$

Then:

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

$y = ln (\frac{{sin}^{2} x}{{cos}^{2} x})$ $y=ln(sin^2x/cos^2x)$

$y = ln ({sin}^{2} x) - ln ({cos}^{2} x)$ $y=ln(sin^2x)-ln(cos^2x)$

$y = 2 ln (sin x) - 2 ln (cos x)$ $y=2ln(sinx)-2ln(cosx)$

Then differentiating becomes easier:

$\frac{d y}{d x} = 2 \cdot \frac{1}{sin x} \cdot \frac{d}{d x} sin x - 2 \cdot \frac{1}{cos x} \cdot \frac{d}{d x} cos x$ $dy/dx=2* 1/sinx * d/dxsinx-2 * 1/cosx * d/dxcosx$

$\frac{d y}{d x} = 2 (\frac{cos x}{sin x} + \frac{sin x}{cos x})$ $dy/dx=2(cosx/sinx+sinx/cosx)$

$\frac{d y}{d x} = \frac{2 ({cos}^{2} x + {sin}^{2} x)}{sin x cos x}$ $dy/dx=(2(cos^2x+sin^2x))/(sinxcosx)$

$\frac{d y}{d x} = \frac{2}{sin x cos x}$ $dy/dx=2/(sinxcosx)$

$\frac{d y}{d x} = 2 sec x csc x$ $dy/dx=2secxcscx$

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How do you find the derivative of $ln (({tan}^{2}) x)$ $ln((tan^2)x)$ ?

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