How do you find dy/dx given $y = ln (2 + x^{2})$ $y=ln(2+x^2)$ ?

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How do you find dy/dx given $y = ln (2 + x^{2})$ $y=ln(2+x^2)$ ?

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Answer 1 · 2016-05-12T20:57:18

$\frac{2 x}{2 + x^{2}}$ $(2x)/(2+x^2)$

Explanation:

Using $\frac{d}{d x} (ln (f (x))) = \frac{1}{f (x)}$ $d/dx(ln(f(x)))=1/f(x)$

combined with the $chain rule$ $color(blue)" chain rule"$

$d/dx[f(g(x))]=f'(g(x)).g'(x)color(green)" A"$
$"------------------------------------------------"$

here $f(g(x))=ln(2+x^2)rArrf'(g(x))=1/(2+x^2)$

and $g(x)=2+x^2rArrg'(x)=2x$
$"---------------------------------------------------------------"$
Substitute these values into $color(green)" A"$

$rArrdy/dx=1/(2+x^2) xx2x=(2x)/(2+x^2)$

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How do you find dy/dx given $y = ln (2 + x^{2})$ $y=ln(2+x^2)$ ?

1 Answer

Explanation:

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