How do you differentiate #g(x) =e^(1-x)coshx# using the product rule?
1 Answer
Jan 8, 2016
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:
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,
This gives:
#g'(x)=-e^(1-x)coshx+e^(1-x)sinhx#
Factored:
#g'(x)=e^(1-x)(sinhx-coshx)#