How do you find the derivative of $y = e^{cosh (2 x)}$ $y = e^cosh(2x)$ ?

Question

How do you find the derivative of $y = e^{cosh (2 x)}$ $y = e^cosh(2x)$ ?

1 Answer

Impact of this question

9067 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-07T11:09:23

$\frac{d}{d x} (e^{cosh (2 x)}) = e^{cosh (2 x)} sinh (2 x) 2$ $frac{d}{dx}(e^{cosh (2x)})=e^{cosh (2x)}sinh (2x)2$

Explanation:

$\frac{d}{d x} (e^{cosh (2 x)})$ $frac{d}{dx}(e^{cosh (2x)})$
Applying the chain rule, $\frac{d f (u)}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x}$ $frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}$
$L e t, cosh (2 x) = u$ $Let,cosh (2x)=u$
$= \frac{d}{d u} (e^{u}) \frac{d}{d x} (cosh (2 x))$ $=frac{d}{du}(e^u)frac{d}{dx}(cosh (2x))$
We know,
$\frac{d}{d u} (e^{u}) = e^{u}$ $frac{d}{du}(e^u)=e^u$
$\frac{d}{d x} (cosh (2 x)) = sinh (2 x) 2$ $frac{d}{dx}(cosh (2x))=sinh (2x)2$
So,
$\frac{d}{d x} (cosh (2 x)) = sinh (2 x) 2$ $frac{d}{dx}(cosh (2x))=sinh (2x)2$

substituted back, $u = cosh (2 x)$ $u=cosh (2x)$

we get,

$e^{cosh (2 x)} sinh (2 x) 2$ $e^{cosh (2x)}sinh (2x)2$

Science

Math

Humanities

... and beyond

How do you find the derivative of $y = e^{cosh (2 x)}$ $y = e^cosh(2x)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the derivative of y = e^cosh(2x)y=ecosh(2x)y = e^cosh(2x)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the derivative of $y = e^{cosh (2 x)}$ $y = e^cosh(2x)$ ?