1. Chain Rule Problems???

1. Find the equation of the tangent line to the curve
at the point .

2. Find the equation of the tangent line to the curve at the point .

3. Differentiate

4. Differentiate

2. 1) $y'=\left\{\left(3\sin x+2\cos x\right)^{-1}\right\}'=-\left(3\sin x+2\cos x\right)^{-2}\left(3\sin x+2\cos x\right)'$ $=-\left(3\sin x+2\cos x\right)^{-2}\left(3\cos x-2\sin x\right)$

2) $y'=5\sec^2 x$

3) $f'(x)=\sec^2 x\left(2\sin x+3\cos x\right)+\tan x\left(2\cos x-3\sin x\right)$

4) $f'(x)=2x\cos x -x^2\sin x$

For the tangent lines use $\ell: \ \frac{y-y_1}{x-x_1}=\frac{dy}{dx}$.

3. thanks alot but how do i write the tangent line equation for number 1 and 2? y=?x+?

4. Use the formula $y - b = m(x - a)$
where m = gradient, a = x co-ordinate of point on line and b = y co-ordinate of point on line

5. Originally Posted by Amanda H
Use the formula $y - b = m(x - a)$
where m = gradient, a = x co-ordinate of point on line and b = y co-ordinate of point on line
Problem 1: y-5=m(x-pi/4)
How do i find m?

6. Originally Posted by Kayla_N
Problem 1: y-5=m(x-pi/4)
How do i find m?
Problem 2 you mean ....?

To get m, differentiate the given function and evaluate the derivative at x = pi/4 (which is why post #2 stated the derivatives for you, in case you were wondering ....)

7. Oh ok..thanks

So is it m= 5????

8. Originally Posted by Kayla_N
Oh ok..thanks

So is it m= 5????
No.

Do you know the exact value of $\cos \frac{\pi}{4}$ and hence the exact value of $\sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} ?$

9. Originally Posted by mr fantastic
No.

Do you know the exact value of $\cos \frac{\pi}{4}$ and hence the exact value of $\sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} ?$

ok so 1/cos(pi/4) = 1.414. then i plug it in to tangent line equation and i got y= 1.14x-6.111 ?? is this right?

10. Originally Posted by Kayla_N
ok so 1/cos(pi/4) = 1.414. [snip]

Google for the exact value of cos(pi/4) if you don't know it and can't look it up.

Originally Posted by Kayla_N
ok so 1/cos(pi/4) = 1.414. then i plug it in to tangent line equation and i got y= 1.14x-6.111 ?? is this right?

What about the squaring and multiplying by 5 to get it from the derivative?? Go back and look at what the derivative is, the one that someone went to the trouble of giving you all those posts ago.

11. ok so cos (pi/4)= sqrt(2)/2. I then square it and time it by 5 and got = 5/2. so now i use the y-y1= m(x-x1) equation???

12. Originally Posted by Kayla_N
ok so cos (pi/4)= sqrt(2)/2. I then square it and time it by 5 and got = 5/2. so now i use the y-y1= m(x-x1) equation???
*Sigh* Please post what the derivative is (and in doing so you will hopefully see why I'm sighing).

13. omg this is so confusing..nevermind then. Thanks for your help.

14. Originally Posted by Kayla_N
omg this is so confusing..nevermind then. Thanks for your help.
No. It's not confusing. You're just not concentrating. Were you not told that $y' = 5 \sec^2 x ?$ It is very simple. You substitute x = pi/4 into this and get the exact value. That gives you the exact value of m:

$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \Rightarrow \sec \frac{\pi}{4} = \frac{2}{\sqrt{2}} \Rightarrow \sec^2 \frac{\pi}{4} = \frac{4}{2} = 2 \Rightarrow m = 5 \sec^2 \frac{\pi}{4} = 5 (2) = 10$.

There is nothing confusing here. It simply requires a small amount of concentration and attention to detail on your part.

Now do you see why I asked you to go back and type the derivative you were given?

15. Wow...gosh I am so dumb...thanks. I should pay more attention next time and read the question carefully. Thanks again for your help..also everyone who helped me.