# Thread: How do I differentiate this?

1. ## How do I differentiate this?

f(x) = 3/quarticroot(9x^2+4)

quarticroot = the square root symbol but with a small 4 on the left side of the root.

2. Think of the quartic root as, instead, being a 1/4 power (square roots are 1/2 power, cube roots are 1/3 power, etc.) So now your expression becomes:

3/(9x^2+4)^(1/4).

You can move your denominator into the numerator simply by making the exponent negative, giving:

3*(9x^2+4)^(-1/4).

Now, you can differentiate this using the power rule in combination with the chain rule. Bring the power down and reduce it by 1 - this makes your new exponent (-5/4). Also, the chain rule says you have to take the derivative of "the inside function," here the 9x^2+4 portion.

This gives 3*(-1/4)(9x^2+4)^(-5/4)*(18x).

Reading left to right, the 3 is the original constant you had;
the (-1/4) is the power you brought down via the power rule;
the (9x^2+4)^(-5/4) is basically your original fourth-root, put in the numerator, with the power reduced by 1;
the 18x is the derivative of 9x^2+4, present because of the chain rule.

You can simplify a little bit, of course, but that takes care of the differentiation.

Hope this helps!

3. Originally Posted by Jeavus
f(x) = 3/quarticroot(9x^2+4)

quarticroot = the square root symbol but with a small 4 on the left side of the root.

$\displaystyle f(x)=\frac{3}{^4\sqrt{9x^2+4}} \Rightarrow 3(9x^2+4)^{\frac{-1}{4}}$
$\displaystyle f'(x)=\frac{3}{4}(9x^2+4)^{\frac{-5}{4}}(18x)$