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

1 Answer
Jan 16, 2017

Answer:

#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))#