# Thread: derivatives + tangent line equation

1. ## derivatives + tangent line equation

For each of the following, compute $\displaystyle \frac{dy}{dx}$ and find the equation for the tangent line at the indicated point x0. You may leave your answers in point-slope form

1. $\displaystyle y = x^2 + cos(pi(x^2+1))$; x0 = 1

for #1.

I found $\displaystyle \frac{dy}{dx}$ and used the chain rule for the cosine part and ended up with
$\displaystyle 2x$

do i then use the tangent line formula ?
$\displaystyle y-y1 = m (x-x1)$?

if so, how do i plug in the numbers?

2. you should have ended up with y' = 2x - sin(pi(x^2 + 1)) (2x * pi)

so for $\displaystyle x_0 = 1$ you have the slope of he point (1,2) to be m=2 .... now you can find the pint slope form equation

3. Originally Posted by Arturo_026
you should have ended up with y' = 2x - sin(pi(x^2 + 1)) (2x * pi)

so for $\displaystyle x_0 = 1$ you have the slope of he point (1,2) to be m=2 .... now you can find the pint slope form equation
am i not supposed to use the chain rule twice for the (pi(x^2+1)
this is what i did,

$\displaystyle y' = 2x - sin(pi(x^2+1)) * (pi(x^2+1))' * (x^2+1)'$
$\displaystyle y' = 2x - sin(pi(x^2+1)) * (0(x^2+1)) * 2x$
$\displaystyle y' = 2x - sin(pi(x^2+1)) * 0 * 2x$
$\displaystyle y' = 2x - 0$
$\displaystyle y' = 2x$

4. Originally Posted by break
am i not supposed to use the chain rule twice for the (pi(x^2+1)
this is what i did,

$\displaystyle y' = 2x - sin(pi(x^2+1)) * (pi(x^2+1))' * (x^2+1)'$
$\displaystyle y' = 2x - sin(pi(x^2+1)) * (0(x^2+1)) * 2x$
$\displaystyle y' = 2x - sin(pi(x^2+1)) * 0 * 2x$
$\displaystyle y' = 2x - 0$
$\displaystyle y' = 2x$
Don't forget that pi is a number:

for example if you are to find the derivative of 3(x^2 + 1) you don't use the chain rule like this: [3(x^2 + 1)]' * 2x , you see?

you just take the 3 out of the differentiation and have: 3 * (x^2 + 1)' ; which gives you 6x

5. Originally Posted by Arturo_026
Don't forget that pi is a number:

for example if you are to find the derivative of 3(x^2 + 1) you don't use the chain rule like this: [3(x^2 + 1)]' * 2x , you see?

you just take the 3 out of the differentiation and have: 3 * (x^2 + 1)' ; which gives you 6x
oh okay, i know pi is a number, but dont i have to find the derivative of that too? since $\displaystyle [pi(x^2+1)]'$ looks like product rule, then the derivative of pi = 0, and anything times zero is zero. why don't i find the derivative of pi?

6. In any case, to avoid confusion why don't just multiply pi times (x + 1) and you'll see that (pi(x+1))' is the same as (x*pi)' + pi'