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Find $y'$ if $y=\sec^{-1}(\sqrt{3x})$. I know that $\sec^{-1}(x) = {1\over x\sqrt{x^2-1}}$ but if I plug in $\sqrt{3x}$ do I use the product rule on the bottom along with the chain rule?

I realize there are easier ways, but I must use Implicit differentiation to solve this problem.

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First, this has nothing to do with implicit differentiation; it’s just an ordinary derivative. Secondly, $\sec^{-1}x$ is most definitely not $\frac1{x\sqrt{x^2-1}}$; what you mean is that the derivative of $\sec^{-1}x$ with respect to $x$ is $\frac1{x\sqrt{x^2-1}}$.

Now let’s look at the actual differentiation problem. You have $f(x)=\sec^{-1}\sqrt{3x}$. Differentiating this will require the use of the chain rule, but not the product rule:

$\begin{align*} \left[\sec^{-1}\sqrt{3x}\right]'&=\frac1{\sqrt{3x}\sqrt{\left(\sqrt{3x}\right)^2-1}}\cdot\left[\sqrt{3x}\right]'\\ &=\frac1{\sqrt{3x}\sqrt{3x-1}}\cdot\left[(3x)^{1/2}\right]'\\ &=\frac1{\sqrt{3x}\sqrt{3x-1}}\cdot\frac12(3x)^{-1/2}\cdot[3x]'\\ &=\frac1{\sqrt{3x}\sqrt{3x-1}}\cdot\frac12\cdot\frac1{\sqrt{3x}}\cdot3\\ &=\frac3{2\sqrt{3x}\sqrt{3x}\sqrt{3x-1}}\\ &=\frac3{6x\sqrt{3x-1}}\\ &=\frac1{2x\sqrt{3x-1}} \end{align*}$

You’re never differentiating a product, so you never use the product rule. (Well, you could use it to differentiate $3x$, but that would be making a great deal of unnecessary work for yourself.)

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It might be easier to proceed as follows.

$\sec(y) = \sqrt{3x} \implies \sec(y) \tan(y) \dfrac{dy}{dx} = \dfrac{\sqrt{3}}{2\sqrt{x}} $

Now plug in for $y$ in terms of $x$, if needed.

Hence, $\dfrac{dy}{dx} = \dfrac{\sqrt{3}}{2\sqrt{x}} \dfrac1{\sec(y) \tan(y)} = \dfrac{\sqrt{3}}{2 \sqrt{x} \sec(y) \tan(y)} = \dfrac{\sqrt{3}}{2 \sqrt{x} \sqrt{3x} \sqrt{\sec^2(y) - 1}} = \dfrac1{2 x \sqrt{3x - 1}}$

For a more "direct" way, using the chain rule, let $z = \sqrt{3x}$, then $y = \sec^{-1}(z)$. Then $\dfrac{dy}{dx} = \dfrac{dy}{dz} \dfrac{dz}{dx} = \dfrac1{z\sqrt{z^2-1}} \dfrac{\sqrt{3}}{2\sqrt{x}}$ Now plug in $z = \sqrt{3x}$ to get the same answer.

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    Please note that I have not learned this method yet. And yes, by using the quotient rule, this would make the problem a lot easier.2012-10-16