Question

How to solve this derivative? with steps

`\frac(1)(e^x + \sqrt {e^x + 1}`

Answer 1

`-(e^x+sqrt(e^x+1))^(-2)(e^x+e^x/(2*sqrt(e^x+1)))`

Explanation:

Writing your term in the form

`(e^x+sqrt(e^x+1))^(-1)` so we get after the rule

`(x^n)'=nx^(n-1)`
`-(e^x+sqrt(e^x+1))^(-2)*(e^x+e^x/(2sqrt(e^x+1)))`
the second factor Comes from the chain rule, if we
differentiate `(e^x+sqrt(e^x+1))` with respect to `x`

Answer 2

`-((e^x + e^x/(2 * sqrt(e^x + 1)))/(e^(2x) + 2(e^x * sqrt(e^x + 1)) + (e^x + 1)))`

Explanation:

Take it in small steps. break the problem down into pieces, and solve those, then put everything together

Start with the formula for finding the derivative of the quotient of 2 functions:

`d/(dx) (f(x))/g(x) = (f'(x) * g(x) - f(x) * g'(x))/g(x)^2`

...this problem allows at least some simplification since the numerator is the constant 1, so `f'(x) = 0`

So, if we denote the denominator in the original function as g(x), the derivative is:

`(0 - g'(x))/(g(x)^2)` (call this equation 1)

We have to then find `g'(x) = d/(dx)(e^x + sqrt(e^x + 1))`

We know that `d/(dx)e^x = e^x`, but we have to find `d/(dx)sqrt(e^x + 1)`

...rewrite the square root term as `(e^x + 1)^(1/2)`
...you find this using the chain rule, and the rule for finding derivatives of powers of functions:

`d/(dx) (f(x)^n) = nf(x)^(n-1)*f'(x)`

so `d/(dx) (e^x + 1)^(1/2) = ((1/2)(e^x + 1)^(-1/2)) * e^x`

so, going back to `g(x) = e^x + sqrt(e^x + 1)`, then

`g'(x) = e^x + ((1/2)(e^x + 1)^(-1/2)) * e^x`

...you can rewrite this as:

`g'(x) = e^x + (e^x/(2 * sqrt(e^x + 1)))`

and now, all we have to do is find `(e^x + sqrt(e^x + 1))^2`:

...and that's just a simple binomial multiplication:

`= (e^(2x) + 2(e^x * sqrt(e^x + 1)) + (e^x + 1))`

So now, you can put all this together:

`(0 - g'(x))/(g(x)^2) = -((e^x + e^x/(2 * sqrt(e^x + 1)))/(e^(2x) + 2(e^x * sqrt(e^x + 1)) + (e^x + 1)))`

GOOD LUCK