How do you differentiate g(x) =x^3 sqrt(4-x) using the product rule?

Apr 23, 2018

$g ' \left(x\right) = 3 {x}^{2} \sqrt{4 - x} - \frac{{x}^{3}}{2 \sqrt{4 - x}}$

Explanation:

$\text{Given "y=f(x)h(x)" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right) h ' \left(x\right) + h \left(x\right) f ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$f \left(x\right) = {x}^{3} \Rightarrow f ' \left(x\right) = 3 {x}^{2}$

$h \left(x\right) = \sqrt{4 - x} = {\left(4 - x\right)}^{\frac{1}{2}}$

$\text{differentiate using the "color(blue)"chain rule}$

$\Rightarrow h ' \left(x\right) = \frac{1}{2} {\left(4 - x\right)}^{- \frac{1}{2}} \times \frac{d}{\mathrm{dx}} \left(4 - x\right)$

$\textcolor{w h i t e}{\Rightarrow h ' \left(x\right)} = - \frac{1}{2 \sqrt{4 - x}}$

$\Rightarrow g ' \left(x\right) = - \frac{{x}^{3}}{2 \sqrt{4 - x}} + 3 {x}^{2} \sqrt{4 - x}$

Apr 23, 2018

$\implies g ' \left(x\right) = \frac{\left(24 - 7 x\right) {x}^{2}}{2 \sqrt{4 - x}}$

Explanation:

We are given:

$g \left(x\right) = {x}^{3} \sqrt{4 - x}$

We apply the product rule as follows:

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] \sqrt{4 - x} + {x}^{3} \frac{d}{\mathrm{dx}} \left[\sqrt{4 - x}\right]$

Simplifying:

$g ' \left(x\right) = \left[3 {x}^{2}\right] \sqrt{4 - x} + {x}^{3} \left[- \frac{1}{2} \frac{1}{\sqrt{4 - x}}\right]$

$g ' \left(x\right) = 3 {x}^{2} \sqrt{4 - x} - \frac{{x}^{3}}{2 \sqrt{4 - x}}$

$g ' \left(x\right) = 3 {x}^{2} \sqrt{4 - x} \cdot \frac{2 \sqrt{4 - x}}{2 \sqrt{4 - x}} - \frac{{x}^{3}}{2 \sqrt{4 - x}}$

$g ' \left(x\right) = \frac{6 {x}^{2} \left(4 - x\right)}{2 \sqrt{4 - x}} - \frac{{x}^{3}}{2 \sqrt{4 - x}}$

$g ' \left(x\right) = \frac{24 {x}^{2} - 6 {x}^{3} - {x}^{3}}{2 \sqrt{4 - x}}$

$g ' \left(x\right) = \frac{24 {x}^{2} - 7 {x}^{3}}{2 \sqrt{4 - x}}$

$\implies \textcolor{b l u e}{g ' \left(x\right) = \frac{\left(24 - 7 x\right) {x}^{2}}{2 \sqrt{4 - x}}}$

Apr 23, 2018

g'(x) = 3x^2sqrt(4-x) - x^3/(2sqrt(4-x)

Explanation:

We can use the product rule with the form;

$g ' \left(x\right) = u ' v + v ' u$

Where we let one term equal to $u$ and the other equal to $v$.
Here;

Let color(blue)(u = x^3

$u$ would have to be differentiated by multiplying the term by the initial power and decreasing the power by one;

$u ' = 3 \cdot {x}^{3 - 1}$
$u ' = 3 {x}^{2}$

And we can do the same for $v$;

Let color(blue)(v = sqrt(4-x)

Just like $u$, $v$ would also have to be differentiated. We can start by changing the square root to the power of $\frac{1}{2}$ to make it easier;

$v = {\left(4 - x\right)}^{\frac{1}{2}}$

By using chain rule;

$v ' = \frac{1}{2} \cdot {\left(4 - x\right)}^{\frac{1}{2} - 1} \cdot \left(4 - x\right) '$

$v ' = - \frac{1}{2} \cdot {\left(4 - x\right)}^{- \frac{1}{2}}$

$v ' = - \frac{1}{2} \cdot \frac{1}{\sqrt{4 - x}}$

color(blue)(v' = -1/(2sqrt(4-x))

Now we can substitute these equations in blue into the equation from the start;

$g ' \left(x\right) = u ' v + v ' u$

$g ' \left(x\right) = 3 {x}^{2} \cdot \sqrt{4 - x} - \frac{1}{2 \sqrt{4 - x}} \cdot {x}^{3}$

g'(x) = 3x^2sqrt(4-x) - x^3/(2sqrt(4-x)