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

Answer 1

William Abdalla

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

the product rule

#color(blue)(y=uv=>(dy)/(dx)=v(du)/(dx)+u(dv)/(dx))#

we have

#g(x)=e^(1-x)sinhx#

#g'(x)=d/(dx)(e^(1-x)sinhx)#

#g'(x)=sinhxd/(dx)(e^(1-x))+(e^(1-x))d/(dx)(sinhx)#

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

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

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Answer 2

Isaiah Allen

To differentiate g(x) = e^(1-x)sinh(x) using the product rule:

Identify the functions within the expression: f(x) = e^(1-x) and h(x) = sinh(x).
Apply the product rule: g'(x) = f'(x)h(x) + f(x)h'(x).
Calculate the derivatives of f(x) and h(x): f'(x) = -e^(1-x) and h'(x) = cosh(x).
Substitute the derivatives and the original functions into the product rule equation: g'(x) = (-e^(1-x))sinh(x) + (e^(1-x))cosh(x).

Therefore, the derivative of g(x) = e^(1-x)sinh(x) using the product rule is g'(x) = (-e^(1-x))sinh(x) + (e^(1-x))cosh(x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.