How to solve this derivative? with steps

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

2 Answers
Jun 4, 2018

-(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

Jun 4, 2018

-((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