## onegirl Group Title Find the derivative of f(x) = cos x, a = pi/2 one year ago one year ago

1. onegirl

@zepdrix i ddi it but can u check my answer?

2. zepdrix

sure :O what are you suppose to do? Just find the derivative then plug $$\large a$$ in? Or does this have something to do with approximation?

3. onegirl

okay my teacher said there is an equation, so here is what i did f(x) = cos (x) f'(x) = -sin(x) x = a = pi/2 so f'(pi/2) = -sin (pi/2) = -1

4. zepdrix

Your teacher said there is an equation? :) That's not much detail lolol Your derivative looks correct at least.

5. onegirl

yea she said try again there is an equation

6. onegirl

let me show you how i wrote it first okay?

7. onegirl

i mean the answer i wrote first

8. zepdrix

k

9. onegirl

d/dx ( cos (x)) = -sin(x) -sin (pi/2) = -1, so the slope is -1, y = mx + b --> y = =1x + b cos (pi/2) = 0 so the coordinates pare are (pi/2, 0) so y = 0x + b, 0(pi/2) + b equals to 0, 0 = y = 0x + 0 = y = 0 for my final answer.

10. zepdrix

Oh so like the previous problems, you're looking for an equation of the tangent line?

11. onegirl

yes sorry i made a mistake

12. onegirl

yes yes the equation of the tangent line

13. zepdrix

$\large y=-x+b$ From here you plugged in your point $$\large \left(\pi/2,\;0\right)$$. $\large 0=-\pi/2+b \qquad \qquad \rightarrow \qquad \qquad b=\pi/2$ I think you plugged your $$\large 0$$ into the wrong spot.

14. onegirl

ohhh okay

15. onegirl

so instead of 0 = 0(pi/2) + b what would i write

16. zepdrix

$\huge (\color{royalblue}{\pi/2},\;\color{orangered}{0}) \qquad = \qquad (\color{royalblue}{x},\;\color{orangered}{y})$ Plug them into here, match up the colors.$\huge \color{orangered}{y}=(-1)\color{royalblue}{x}+b$ I'm not sure why you're plugging a $$\large 0$$ in for your -1....

17. onegirl

okay

18. onegirl

so 0 = (-1)(pi/2) + b ?

19. zepdrix

yes

20. onegirl

okay

21. onegirl

thx

22. zepdrix

np