# Derivative sinhx*tanhx

#### MechEng

When I calculate the derivative of

$$\displaystyle y = sinh(x) tanh(x)$$

I come up with

$$\displaystyle y' = cosh(x) tanh(x) + sinh(x) sech^2(x)$$

Should this be further simplified?

#### General

$$\displaystyle cosh(x) \cdot tanh(x)=sinh(x)$$

then take $$\displaystyle sinh(x)$$ as a common factor ..

Unknown008

#### MechEng

So...

$$\displaystyle y' = sinh(x)(1+sech^2(x))$$

Sorry, it has been a very long time since I have had to work these sorts of problems.

Correct .