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

1 Answer
Jan 8, 2016

Answer:

#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)#