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Answer 1 · 2017-07-30T22:43:08

$d/(dx)[(sqrt(x+1))/x^2]=1/(2x^2sqrt(x+1))-(2sqrt(x+1))/x^3$

$[(f(x))/(g(x))]'=(f'(x)g(x)-f(x)g'(x))/(g(x)^2)$

In this case

$f(x)=sqrt(x+1)$

and

$g(x)=x^2$

To find $f'(x)$ we will use the Chain Rule, which states

$[h(k(x))]'=h'(k(x))k'(x)$

In this case

$h(x)=sqrt(x)$

and

$k(x)=x+1$

Since the derivative of a constant is $0$ and the derivative of $x$ is $1$

$k'(x)=1$

Also remember that $sqrt(x)=x^(1/2)$

and the Power rule tells us that $(x^n)'=nx^(n-1)$

Then

$h'(x)=1/2x^(1/2-1)=1/2x^(-1/2)=1/(2sqrt(x)$

Then we plug in and find that

$f'(x)=1/(2sqrt(x+1))$

and we can use the power rule again to find $g'(x)$

$g'(x)=(x^2)'=2x^(2-1)=2x^1=2x$

Then we plug in

$(sqrt(x+1)/(x^2))'=((x^2)/(2sqrt(x+1))-2xsqrt(x+1))/(x^2)^2$

and we simplify

$=((x^2)/(2sqrt(x+1))-2xsqrt(x+1))/(x^4)$

$=((x^2)/(2sqrt(x+1)))x^-4-(2xsqrt(x+1))/x^4$

$=x^-2/(2sqrt(x+1))-(2sqrt(x+1))/x^3$

$=1/(2x^2sqrt(x+1))-(2sqrt(x+1))/x^3$