Thread: Finding slope of a tangent to a function

Hi guys,

Another question I'm confused with. Question is:

Find the slope of the tangent to the function f(x)=2^x^2+3x when x=3.

First I start with the chain rule:

f(x)=2^(x^2+3x)
f'(x)=2^(x^2+3x) ln2 (2x+3)

Then I input the value of x=3 into the equation.

f'(3)=2^(3^2+3(3)) ln2 (2(3)+3)
f'(3)=2^(3^2+3(3)) ln2 (9)

This is where I get stuck, if I complete the function I am left with:

f'(3)=2^18ln2 (9)
f'(3)= 1,635,229.37. Obviously this is not the correct answer.

I'm stuck, again.

Originally Posted by Butum

Hi guys,

Another question I'm confused with. Question is:

Find the slope of the tangent to the function f(x)=2^x^2+3x when x=3.

First I start with the chain rule:

f(x)=2^x^2+3x
f'(x)=2^x^2+3x ln2 (2x+3)

Then I input the value of x=3 into the equation.

f'(3)=2^3^2+3(3) ln2 (2(3)+3)
f'(3)=2^3^2+3(3) ln2 (9)

This is where I get stuck, if I complete the function I am left with:

f'(3)=2^18ln2 (9)
f'(3)= 1,635,229.37. Obviously this is not the correct answer.

I'm stuck, again.

first, these questions should be posted in the calculus forum, not pre-calculus.

second, your notation leaves much to be desired. use parentheses to make your expressions clear.

from your work, I assume your function is

f(x) = 2^(x^2+3x) rather than f(x) = 2^(x^2) + 3x

f'(x) = 2^(x^2+3x) * (2x+3) * ln(2)

f'(3) = 2^(18) * (9) *ln(2) ... which, yes, is a very large number.

You are correct with your assumptions, I've edited the original post.

Can this thread be moved to the proper section?

I still have not been able to solve this problem, perhaps I'm using the wrong rule?

Originally Posted by Butum

I still have not been able to solve this problem, perhaps I'm using the wrong rule?

Thanks skeeter, that's one of the two answers that I have been able to come up with.

Is it common to see numbers that large when using the chain rule on polynomial expressions in the exponent? At first glance I feel like that is a very wrong answer.