1. ## Differentiation Hyperbolic Function

Differentiate $y=\mathrm{sinh}^{n-1}x\mathrm{cosh}x$ with respect to $x$.

2. Originally Posted by Air
Differentiate $y=\mathrm{sinh}^{n-1}x\mathrm{cosh}x$ with respect to $x$.

$y'=\sinh^{n-1}(x)\cdot\left(\cosh(x))'+\cosh(x)\cdot(\sinh^{n-1}(x)\right)'$

$\left(\cosh(x)\right)'=\left(\frac{e^x+e^{-x}}{2}\right)'=\frac{e^x-e^{-x}}{2}=\sinh(x)$

and $\left(\sinh^{n-1}(x)\right)'=(n-1)\left(\sinh^{n-2}(x)\right)\cdot\left(\frac{e^x-e^{-x}}{2}\right)'=$ $(n-1)\sinh^{n-2}(x)\cdot\frac{e^x+e^{-x}}{2}=(n-1)\sinh^{n-2}(x)\cdot\cosh(x)$

3. Hello, Air!

Differentiate $y\:=\: \sinh^{n-1}\!x\cdot \cosh x$

Product Rule: . $y' \;=\;\sinh^{n-1}\!x\cdot\sinh x + (n-1)\sinh^{n-2}\!x\cdot\cosh x$

. . . . . . . . . . $y' \;=\;\sinh^n\!x + (n-1)\sinh^{n-2}\!x\cdot\cosh x$

4. Originally Posted by Soroban
Hello, Air!

Product Rule: . $y' \;=\;\sinh^{n-1}\!x\cdot\sinh x + (n-1)\sinh^{n-2}\!x\cdot\cosh x$

. . . . . . . . . . $y' \;=\;\sinh^n\!x + (n-1)\sinh^{n-2}\!x\cdot\cosh x$

Isn't it: $y' \;=\;\sinh^n\!x + (n-1)\sinh^{n-2}\!x\cdot\cosh ^2 x$?

5. Originally Posted by Air
Isn't it: $y' \;=\;\sinh^n\!x + (n-1)\sinh^{n-2}\!x\cdot\cosh ^2 x$?
Yes

6. Hello, Air!

Isn't it: $y' \;=\;\sinh^n\!x + (n-1)\sinh^{n-2}\!x\cdot\cosh ^2 x$?

Yes . . . Sorry!