Question

The Hyperbolic Functions and their Inverses: For our purposes, the hyperbolic functions, such as sinhx=ex−e−x2andcoshx=ex+e−x2 are simply extensions of the exponential, and any questions concerning them can be answered by using what we know about exponentials. They do have a host of properties that can become useful if you do extensive work in an area that involves hyperbolic functions, but their importance and significance is much more limited than that of exponential functions and logarithms. Let f(x)=sinhxcoshx.
d/dx f(x) =_________

Answer 1

Answer:

$\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}$

Step-by-step explanation:

It is given that

$\sinh x=\dfrac{e^x-e^{-x}}{2}$

$\cosh x=\dfrac{e^x+e^{-x}}{2}$

$f(x)=\sinh x\cosh x=$

Using the given hyperbolic functions, we get

$f(x)=\dfrac{e^x-e^{-x}}{2}\times \dfrac{e^x+e^{-x}}{2}$

$f(x)=\dfrac{(e^x)^2-(e^{-x})^2}{4}$

$f(x)=\dfrac{e^{2x}-e^{-2x}}{4}$

Differentiate both sides with respect to x.

$\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left(\dfrac{e^{2x}-e^{-2x}}{4}\right )$

$\dfrac{d}{dx}f(x)=\left(\dfrac{2e^{2x}-(-2)e^{-2x}}{4}\right )$

$\dfrac{d}{dx}f(x)=\left(\dfrac{2(e^{2x}+e^{-2x})}{4}\right )$

$\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}$

Hence, $\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}$.