Q2) Given that f(x) = (sinh x)/(1+cosh x) then f'(x) = a)...

Q2) Given that $f(x) = \frac{\sinh x}{1+\cosh x}$ then $f'(x) = $\\ $a) \frac{1}{(1+\cosh x)^2}$ $b) \frac{\sinh x}{1+\cosh x}$ $c) \frac{1}{1+\cosh x}$ $d) \frac{\cosh x}{(1+\cosh x)^2}$

Transcript

00:01 It's clear as some enumerate here.

00:04 So we have d of kosh of x over d x is equal to one half d e to the x plus e to the negative x over d x this becomes equal to one half of e to the x minus e to the negative x so i see that d of d over d x of kosh is equal to the hyperbolic function of sign for part b we have the 10 which is equal to sine over cosine and we get e to the x minus e to the negative x all over e to the x plus e to the negative x all over e to the x plus e to the negative x so we're going to differentiate d -y over d x.

01:47 Then we get e to the x minus e to the negative x the derivative e to the x plus e to the negative x minus e to the negative x minus e to the negative x times the derivative of e to the x plus e to the negative x all over e to the x plus e to the negative x squared.

02:26 When we end up getting e to the 2x plus 2 plus e to the negative 2x plus 2 minus e to the negative 2x all over e to the x plus e to the negative x square which is equal to 2 over e to the x plus e to the negative x square then we get 1 over e to the x plus e to the negative x over 2 square which is equal to 1 over the hyperbolic function of cosine square and we get secant square of x for part c we have co -secent is equal to 1 over sign which is equal to 1 over 1 half e to the x minus e to the negative x, which is equal to 2 times e to the x minus e to the negative x, the negative 1 power.

04:34 We differentiate dy over dx, and we get negative 2 times negative 1 times e to the x minus e to the negative x to the negative second power times the derivative of e to the x minus e to the negative x, which is equal to negative 2 over e to the x minus e to the negative x times each the x plus e to the negative x all over e to the x minus negative e to the x, which is equal to negative co -secant times cosine over sign.

05:46 This is equal to negative co -secent times cotangent...

Question

Q2) Given that $f(x) = \frac{\sinh x}{1+\cosh x}$ then $f'(x) = $\\ $a) \frac{1}{(1+\cosh x)^2}$ $b) \frac{\sinh x}{1+\cosh x}$ $c) \frac{1}{1+\cosh x}$ $d) \frac{\cosh x}{(1+\cosh x)^2}$

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