Differentiation - chain and product rule.

[xxChrisxx]

It's been years since I've done maths properly so I'm rusty with it. I'm helping out a colleague at work who is studying maths for an OU course.

Homework Statement

Part 1: Differentiate function.
f(x) = e^(0.5x+cos(x))

Part 2: Use answer from part 1 to show.
g(x) = (1+2sinx)(e^(0.5x+cos(x)))
Has the derivative:
g'(x) = 0.5(4cos^2(x)+4cos(x)-3)(e^(0.5x+cos(x)))

The Attempt at a Solution

A=e^(0.5x+cos(x))

Part 1:
Using chain rule
f'(x) = A(0.5-sin(x))


Part 2:
Using this and the product rule I get.

g'(x) = 0.5A(4cos(x)-4sin^2(x)+1)


Which isn't the same.

 

[HallsofIvy]

It's been years since I've done maths properly so I'm rusty with it. I'm helping out a colleague at work who is studying maths for an OU course.

Homework Statement

Part 1: Differentiate function.
f(x) = e^(0.5x+cos(x))

Part 2: Use answer from part 1 to show.
g(x) = (1+2sinx)(e^(0.5x+cos(x)))
Has the derivative:
g'(x) = 0.5(4cos^2(x)+4cos(x)-3)(e^(0.5x+cos(x)))

The Attempt at a Solution

A=e^(0.5x+cos(x))

Part 1:
Using chain rule
f'(x) = A(0.5-sin(x))


Part 2:
Using this and the product rule I get.

g'(x) = 0.5A(4cos(x)-4sin^2(x)+1)


Which isn't the same.

Yes, it is the same. Use the fact that $sin^2(x)= 1- cos^2(x)$.

 

[xxChrisxx]

Ahh I forgot about that, thanks.