- Thread starter puremath
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\(\displaystyle y=k\cdot f(x)\implies y'=k\cdot f'(x)\)

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

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.

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

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

You should be able to show:

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

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

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.

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

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

You should be able to show:

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

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

I 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.Yes, good work!

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

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

Using W/A, cosh(0) = 1.Since \(\displaystyle \sinh'(x)=\cosh(x)\), yes.

sinh(1) is a transcendental number whose value is (e^2 - 1)/(2e).