How do you find the derivative of f(x)=sqrt(1+3x^2)f(x)=1+3x2?

1 Answer
Jan 16, 2017

f(x)=sqrt(1+3x^2) => f'(x)=(3x)/sqrt(1+3x^2)

Explanation:

First remember sqrt(x)=x^(1/2), then we can write
f(x)=sqrt(1+3x^2)=(1+3x^2)^(1/2)

and remember the chain rule

f(g(x))'=f'(g(x))g'(x)

So, let

h(x)=sqrt(x)

and

g(x)=1+3x^2

Then we can write

f(x)=(h@g)(x)=h(g(x))=sqrt(g(x))

then

f'(g(x))=(sqrt(g(x)))'=(g(x)^(1/2))'

which by the power rule ((x^n)'=nx^(n-1))

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

then we can plug in g(x)

h'(g(x))=1/2(1+3x^2)^(-1/2)

and

g(x)=1+3x^2 => g'(x)=(1)'+(3x^2)'=0+6x=6x

by the fact that the derivative of sums is the sum of derivatives
and the power rule

then since,

f(x)=(h@g)(x) => f'(x)=h'(g(x))g'(x)

we can plug in our results

f'(x)=6x(1/2(1+3x^2)^(-1/2))=3x(1+3x^2)^(-1/2)=underline((3x)/sqrt(1+3x^2))