How do you find the derivative of $f (x) = \sqrt{1 + 3 x^{2}}$ $f(x)=sqrt(1+3x^2)$ ?

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Answer 1 · 2017-01-16T02:27:57

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

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

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

How do you find the derivative of f(x)=sqrt(1+3x^2)f(x)=√1+3x2f(x)=sqrt(1+3x^2)?