How do you differentiate (sqrt x)(x^2+3sinx)?

Feb 11, 2018

Using differentiate product rule
$\frac{1}{2} \left({x}^{2} + 3 \sin x\right) \frac{\mathrm{dx}}{\sqrt{x}} + \sqrt{x} \left(2 x + 3 \cos x\right) \mathrm{dx}$
See explanation below

Explanation:

We know that in functions product, de derivative is (derivative of a product)

(f·g)´dx=f´·g·dx+f·g´·dx

Thus we have (if $f \left(x\right) = \sqrt{x}$ and $g \left(x\right) = {x}^{2} + 3 \sin x$)

f´(x)=1/2 1/(sqrtx)

g´(x)=2x+3cosx.

If we apply prior formula we have

(f·g)´dx=f´·g·dx+f·g´·dx = $\frac{1}{2} \left({x}^{2} + 3 \sin x\right) \frac{\mathrm{dx}}{\sqrt{x}} + \sqrt{x} \left(2 x + 3 \cos x\right) \mathrm{dx}$

Feb 11, 2018

$\frac{{x}^{2} + 3 \sin \left(x\right)}{2 \sqrt{x}} + \sqrt{x} \left(2 x + 3 \cos \left(x\right)\right)$

Explanation:

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

Obviously we already have f(x) and g(x) so we just need to find the derivatives of both

f'(x)=$\frac{d}{\mathrm{dx}} \left[\sqrt{x}\right] = \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{2}}\right] = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$

g'(x)=$\frac{d}{\mathrm{dx}} \left[{x}^{2} + 3 \sin \left(x\right)\right] = 2 x + 3 \cos \left(x\right)$
To derive the 3sin(x) you would have to use product rule again where f(x)=3 and g(x)=sin(x), but remember the derivative of a plain number is zero, so all that's left of product rule is #f(x)*g'(x), which is just 3cos(x),remember the derivative of sin is cos.

Now just plug all the parts yo have into product rule to get:$\frac{{x}^{2} + 3 \sin \left(x\right)}{2 \sqrt{x}} + \sqrt{x} \left(2 x + 3 \cos \left(x\right)\right)$