PracticeFind S'(x) & C'(x)

pre-trip_rapture

Member
I know the quotient rule is used to find S'(x) and C'(x). Find S'(x) and C'(x). I will then find the following: S'(0) and C'(1).

MarkFL

La Villa Strangiato
Staff member
Moderator
Math Helper
We don't need the quotient rule, just the following rules:

$$\displaystyle y=k\cdot f(x)\implies y'=k\cdot f'(x)$$

$$\displaystyle y=e^{f(x)}\implies y'=e^{f(x)}\cdot f'(x)$$

You should be able to show:

$$\displaystyle \frac{d}{dx}\sinh(x)=\cosh(x)$$

$$\displaystyle \frac{d}{dx}\cosh(x)=\sinh(x)$$

pre-trip_rapture

pre-trip_rapture

Member
We don't need the quotient rule, just the following rules:

$$\displaystyle y=k\cdot f(x)\implies y'=k\cdot f'(x)$$

$$\displaystyle y=e^{f(x)}\implies y'=e^{f(x)}\cdot f'(x)$$

You should be able to show:

$$\displaystyle \frac{d}{dx}\sinh(x)=\cosh(x)$$

$$\displaystyle \frac{d}{dx}\cosh(x)=\sinh(x)$$
I surely will follow-up on this problem. I want to do the math and then evaluate the derivative for each function. I am not in calculus yet but there are derivatives and some integration questions that I can do even now. Going to bed.

P. S. I know you hate me for saying this will be done at a later time but I want to do the work, right or wrong as a follow-up.

pre-trip_rapture

Member
We don't need the quotient rule, just the following rules:

$$\displaystyle y=k\cdot f(x)\implies y'=k\cdot f'(x)$$

$$\displaystyle y=e^{f(x)}\implies y'=e^{f(x)}\cdot f'(x)$$

You should be able to show:

$$\displaystyle \frac{d}{dx}\sinh(x)=\cosh(x)$$

$$\displaystyle \frac{d}{dx}\cosh(x)=\sinh(x)$$

Staff member
Moderator
Math Helper
Yes, good work!

pre-trip_rapture

Member
Yes, good work!
I know just the basics of Calculus 1 and 2, just the basics. When we get there after Cohen, I want to dive deep into calculus, maybe even 4 or 5 limits using delta and epsilon. There is a sleep over event at the job. I requested tonight off two weeks ago.

pre-trip_rapture

Member
Yes, good work!
To find S'(0), I must solve cosh(0), right?

To find C'(1), I find solve sinh(1), right?

MarkFL

La Villa Strangiato
Staff member
Since $$\displaystyle \sinh'(x)=\cosh(x)$$, yes.
Since $$\displaystyle \sinh'(x)=\cosh(x)$$, yes.