# How do you use the product rule to find the derivative of y=sqrt(x)*cos(x) ?

Aug 6, 2014

The product rule states:

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

So, if we are trying to find the derivative of $y = \sqrt{x} \cdot \cos x$, then we will let $f \left(x\right) = \sqrt{x}$ and $g \left(x\right) = \cos x$.

Then, by the product rule, we have:

$\frac{d}{\mathrm{dx}} \left[\sqrt{x} \cdot \cos x\right] = \frac{d}{\mathrm{dx}} \left[\sqrt{x}\right] \cdot \cos x + \sqrt{x} \cdot \frac{d}{\mathrm{dx}} \left[\cos x\right]$

$\sqrt{x}$ is the same thing as ${x}^{\frac{1}{2}}$. Therefore, by the power rule, the derivative of ${x}^{\frac{1}{2}}$ is $\frac{1}{2} {x}^{- \frac{1}{2}}$. And, we know from basic trig derivative rules that $\frac{d}{\mathrm{dx}} \left[\cos x\right] = - \sin x$.

So, now we will substitute into our little formula:

d/dx[sqrt(x) ⋅ cosx] = 1/2 x^(-1/2)⋅cosx + sqrt(x)⋅(-sin x)

Recalling that ${x}^{- \frac{1}{2}}$ is the same thing as $\frac{1}{\sqrt{x}}$, we will simplify this equation:

d/dx[sqrt(x) ⋅ cosx] = cosx/(2sqrt(x)) - sinx sqrt(x)

And there is our derivative. Remember, when you're differentiating radicals, it's always helpful to rewrite things with rational exponents. That way, you can find derivatives easily using the power rule.