Question

How do you differentiate `g(x) =e^(1-x)coshx` using the product rule?

Answer 1

`g'(x)=e^(1-x)(sinhx-coshx)`

Explanation:

According to the product rule, which will need to be used here,

`g'(x)=coshxd/dx(e^(1-x))+e^(1-x)d/dx(coshx)`

Find each derivative separately.

The first will require the chain rule: `d/dx(e^u)=e^u*u'`

Thus,

`d/dx(e^(1-x))=e^(1-x)*d/dx(1-x)=e^(1-x)(-1)=-e^(1-x)`

The second requires the knowledge of hyperbolic trig functions. One of their basic characteristics is that hyperbolic sine and hyperbolic cosine are the derivatives of one another:

`d/dx(coshx)=sinhx`

(Also, `d/dx(sinhx)=coshx`.)

This gives:

`g'(x)=-e^(1-x)coshx+e^(1-x)sinhx`

Factored:

`g'(x)=e^(1-x)(sinhx-coshx)`