What is the derivative of $y = sin (1 + tan (2 x))$ $y=sin(1+tan(2x))$ ?

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Answer 1 · 2016-08-19T23:56:33

$2 {sec}^{2} (2 x) \cdot cos (1 + tan 2 x)$ $2sec^2(2x)*cos(1+tan2x)$

Finding the derivative of this function by applying chain rule and using derivative of some trigonometric functions.

Let:
$u (x) = 1 + tan (2 x)$ $u(x)=1+tan(2x)$
$f (x) = sin x$ $f(x)=sinx$

Then $f (u (x)) = sin (1 + tan (2 x))$ $f(u(x))=sin(1+tan(2x))$

$f (u (x)))$ $f(u(x)))$ is a composite function where its derivative can be found by applying chain rule that says:

$color(blue)(f(u(x))'=u'(x)*f'(u(x))$

Let's find $u'(x)$ and $f'(x)$

$u'(x)=(2x)'*tan'(2x)$
$color(blue)(u'(x)=2sec^2(2x))$

$color(blue)(f'(x)=cosx)$

$(f(u(x)))'=u'(x)*f'(u(x))$

$(f(u(x)))'=2sec^2(2x)*cos(u(x))$

$color(red)((sin(1+tan2x))'=2sec^2(2x)*cos(1+tan2x))$

What is the derivative of y=sin(1+tan(2x))y=sin(1+tan(2x))y=sin(1+tan(2x))?