Equation of a tangent to a complicated function

[brizer]

The problem: The y-intercept of the line tangent to y=(x⁴-2x²-8)e^cosx2 at x=1 is (a) -15.499 (b) -25.999 (c) -41.448 (d) 15.449 (e) 25.99

To find the y-intercept I have to find the whole equation of the line
y=f(1)
m=f'(1)
x=1
b=??

My problem is that I don't think I got the the derivative of f(x) right.
I know you have to apply the product rule and the chain rule, but I don't know if the chain rule should be applied to e^cosx2. I came up with this:
(4x³-4x)e^cosx2 + (x⁴-2x²-8)e^cosx2(-sinx²)(2x)

then y=-15.4487
m =15.1465
and b=-1.020

the intercept doesn't equal any of the given answers, but y= answer a. However, if I use f'(x)= (4x³-4x)e^cosx2 + (x⁴-2x²-8)e^cosx2, I get b=1. What did I do wrong?

 

[HallsofIvy]

Yes, of course, you need to use the chain rule. But I don't see where you got m. When x= 1, the derivative (slope of tangent line) is (4-4)e^cos(1)+ (1- 2- 8)e^cos(1)(-sin(1))(2)= 18e^cos(1)sin(1) and I do NOT get 15.465 for that!

 

[brizer]

you're my hero. I made a silly mistake. Thanks so much for catching it!