Please help and explains to me how to do it...Thanks
1. Find the equation of the tangent line to the curveat the point .
2. Find the equation of the tangent line to the curve at the point .
3. Differentiate
4. Differentiate
Please help and explains to me how to do it...Thanks
1. Find the equation of the tangent line to the curveat the point .
2. Find the equation of the tangent line to the curve at the point .
3. Differentiate
4. Differentiate
1) $\displaystyle y'=\left\{\left(3\sin x+2\cos x\right)^{-1}\right\}'=-\left(3\sin x+2\cos x\right)^{-2}\left(3\sin x+2\cos x\right)'$ $\displaystyle =-\left(3\sin x+2\cos x\right)^{-2}\left(3\cos x-2\sin x\right)$
2) $\displaystyle y'=5\sec^2 x$
3) $\displaystyle f'(x)=\sec^2 x\left(2\sin x+3\cos x\right)+\tan x\left(2\cos x-3\sin x\right)$
4) $\displaystyle f'(x)=2x\cos x -x^2\sin x$
For the tangent lines use $\displaystyle \ell: \ \frac{y-y_1}{x-x_1}=\frac{dy}{dx}$.
No! ..... exact value please.
Google for the exact value of cos(pi/4) if you don't know it and can't look it up.
Your gradient m is totally wrong.
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.
No. It's not confusing. You're just not concentrating. Were you not told that $\displaystyle y' = 5 \sec^2 x ?$ It is very simple. You substitute x = pi/4 into this and get the exact value. That gives you the exact value of m:
$\displaystyle \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \Rightarrow \sec \frac{\pi}{4} = \frac{2}{\sqrt{2}} \Rightarrow \sec^2 \frac{\pi}{4} = \frac{4}{2} = 2 \Rightarrow m = 5 \sec^2 \frac{\pi}{4} = 5 (2) = 10$.
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?