Question

what is the second derivative of 2^(x+1) +cotX

Answer 1

do you know the first derivative?

Answer 2

I think that the first D is x+1 2^X -csc^2x

Answer 3

\[\large \frac{d}{dx} a^x = a^x \cdot \ln(a)\]

Answer 4

where a is a numerical value raised to a function.

Answer 5

you got the second half of the problem right, \(\large\frac{d}{dx} \cot(x) = -\csc^2(x)\)

Answer 6

so then it would be 2^x=1 ln 2^x+1 -csc^2X

Answer 7

that should be 2^x+1Ln2 -csc^2x

Answer 8

No you're right, I made an error, lol. Good job.

Answer 9

thanks! but the first D is not finished is it?

Answer 10

\[\large f'(x) = 2^{x+1}\cdot \ln(2) -\csc^2(x)\] and you're going to take the derivative of that...

Answer 11

ok I will give it a go

Answer 12

Hint: use product rule.

Answer 13

do I Look at Ln2 as being a coefficient such that the derivative of it would be 0?

Answer 14

Yes.

Answer 15

so then the final answer is (cscx+cotx)^2

Answer 16

This is what I got.

Answer 17

Started by breaking the first derivative in half.

Answer 18

One min.

Answer 19

\[\large f'(x)= 2^{x+1}\cdot \ln(2)-\csc^2(x)\]\[\large f''(x) = [2^{x+1}\cdot \ln(2) \cdot \ln(2)-0]-\csc^2(x)\]\[\large f''(x) = 2^{x+1}\cdot \ln^2(2) - \csc^2(x)\]\[\large \ \ \ \frac{d}{dx} (\csc^2(x)) = -2\csc^2(x)\cot(x) \ (\text{ with u-sub})\]\[\large f''(x) = 2^{x+1} \cdot \ln^2(2) +2\csc^2(x)\cot(x)\]\[\large \ \ \ 2^{x+1} = 2^x\cdot 2^1\]\[\large f''(x)=\color{red}{2[2^x\ln^2(2)+\csc^2(x)\cot(x)]}\]

Answer 20

let me know if there is anything /any step that confuses you. :)

Answer 21

isn't d/dx csc^2x = 2cscx(cscxcotx)^2

Answer 22

Nope,there is a negative in front of it.

Answer 23

Ok thanks! the book doesn't give a whole lot of ex!

Thanks ;)

Answer 24

\[\large \frac{d}{dx} (\csc^2(x))\] take \(u=\csc(x)\) and \(du=-\csc(x)\cot(x)\)\[\large \frac{d}{dx}(u^2)=2udu=2(\csc(x))\cdot (-\csc(x)\cot(x))=-2\csc^2(x)\cot(x)\] :)

Answer 25

Do you understand?

Answer 26

Yes it is coming slowly. But I see what you have done. you took the and used substitution and then the derivative of u^2

Answer 27

Yess :)

Answer 28

thanks again you da best!