Help finding the first AND SECOND derivative of y=cos2x?

Question

How to take 1st AND 2nd derivative of y=cos2x? Meaning take the 1st derivative and then take the derivative of that.

And also

the square root of x^2+1

and

(3s+5)^8

Please explain how to do it!! not just the answers.

Answer 1

Sounds like you slept through a class. If you are taking d(cos^2 x)/dx, you can do this by the chain rule and sneaky substitution. Let u= cos x and du/dx= -sin x. Then d(cos^2)/dx= d (u^2)/du * du/dx. Which is (2 cosx)(-sinx), Then you take the derivitive of this, which uses another rule:
If y=g(x)f(x), dy/dx = g(x)* d(f(x))/dx + d(g(x))/dx*f(x).

To take the derivitive of sqrt(x^2+1), Let u= x^2+1, so du/dx= 2x. Then you have sqrt(u) to take the derivitive of, which is 1/2 (1/sqrt(u))du/dx.
or x/ sqrt(x^2+1).

The last one can be done the same way u=(3s+5) and du/ds=3. So the problem converts d u^8/du^ du/ds or 24 (3s+5)^7

Answer 2

sin(2) is erroneous! it would be cos(2x) * 2 or 2cos(2x). you ought to use the chain rule on the interior and that's the place you get the two. and you utilising the trig. derivatives all of us be attentive to all of us be attentive to the by-made from sine is cosine.

Answer 3

first AND SECOND derivative of y=cos2x
dy/dx = d/dx(cos2x) = -2sin(2x)
d^2y/dx^2 = d^2/dx^2 (-2sin(2x))
= -4cos2x

d/dx (sqrt(x^2 +1) )
= (1/2)/ sqrt(x^2 +1) (2x)
= x / sqrt(x^2 +1)

Answer 4

y'=-2sin2x
y''=-4cos2x

y'=-x/sqrt(x^2+1)

24(3s+5)^7