If fx-tan 2x tan 3x tan 5x- then prove that f-x-5 2 5x -3 2 3x -2 2 2x

tan 5x =tan( 2x+3x ) #160; =tan 2x +tan 3x #160; 1 tan 2x tan 3x #160; ∴ f(x)=tan 5x tan 2x tan 3x ∴ f'(x)=5 ^ 2 5x 3 ^ 2 3x ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

If fx=tan 2...

Question

If $f (x) = tan 2 x tan 3 x tan 5 x$ , then prove that $f^{'} (x) = 5 {sec}^{2} 5 x - 3 {sec}^{2} 3 x - 2 {sec}^{2} 2 x$

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Solution

$tan 5 x = tan (2 x + 3 x) = \frac{tan 2 x + tan 3 x}{1 - tan 2 x tan 3 x}$
$∴ f (x) = tan 5 x - tan 2 x - tan 3 x$
$∴ f^{'} (x) = 5 {sec}^{2} 5 x - 3 {sec}^{2} 3 x - 2 {sec}^{2} 2 x$

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If $F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}$ where $f^{'} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

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Q. If

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and given that

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and

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If f(x)=tan2xtan3xtan5x, then prove that f′(x)=5sec25x−3sec23x−2sec22x

tan5x=tan(2x+3x)=tan2x+tan3x1−tan2xtan3x∴f(x)=tan5x−tan2x−tan3x∴f′(x)=5sec25x−3sec23x−2sec22x

If $f (x) = tan 2 x tan 3 x tan 5 x$ , then prove that $f^{'} (x) = 5 {sec}^{2} 5 x - 3 {sec}^{2} 3 x - 2 {sec}^{2} 2 x$

$tan 5 x = tan (2 x + 3 x) = \frac{tan 2 x + tan 3 x}{1 - tan 2 x tan 3 x}$
$∴ f (x) = tan 5 x - tan 2 x - tan 3 x$
$∴ f^{'} (x) = 5 {sec}^{2} 5 x - 3 {sec}^{2} 3 x - 2 {sec}^{2} 2 x$