Question

How do you find the derivative with a square root in the denominator $y=5\dfrac{x}{\sqrt{{{x}^{2}}+9}}$?

Answer 1

We recall the quotient rule of differentiation that is for some differentiable functions $f\left( x \right),g\left( x \right)$ given as $\dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)\dfrac{d}{dx}f\left( x \right)-f\left( x \right)\dfrac{d}{dx}g\left( x \right)}{{{g}^{2}}\left( x \right)}$. We take $f\left( x \right)=5x,g\left( x \right)=\sqrt{{{x}^{2}}+9}$ and use quotient rules. We find $\dfrac{d}{dx}g\left( x \right)=\dfrac{d}{dx}\sqrt{{{x}^{2}}+9}$ using the chain rule differentiation $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$ where we take $u={{x}^{2}}+9$ and $y=\sqrt{{{x}^{2}}+9}$.$$$$

Complete step by step answer:
We know that the process of finding derivative is called differentiation. We know that if we are given two differentiable functions $f\left( x \right),g\left( x \right)$ in their respective domain with the condition on denominator $g\left( x \right)\ne 0$ then we can differentiate the quotient using the following rule as
$\dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)\dfrac{d}{dx}f\left( x \right)-f\left( x \right)\dfrac{d}{dx}g\left( x \right)}{{{g}^{2}}\left( x \right)}$
If we are asked to differentiate using the composite function $y=g\left( f\left( x \right) \right)$ then we can differentiate using the chain rule as taking $u=g\left( x \right)$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$
We are asked to differentiate the derivative of following with a square root in the denominator
$y=5\dfrac{x}{\sqrt{{{x}^{2}}+9}}=\dfrac{5x}{\sqrt{{{x}^{2}}+9}}$
We take the numerator function $f\left( x \right)=5x$ and $g\left( x \right)=\sqrt{{{x}^{2}}+9}$ . We first find differentiate $f\left( x \right)=5x$ with respect to $x$ to have;
$\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}5x=5\dfrac{d}{dx}x=5\cdot 1=5$
We first find differentiate $g\left( x \right)=\sqrt{{{x}^{2}}+9}$ with respect to $x$ to have;
$\dfrac{d}{dx}g\left( x \right)=\dfrac{d}{dx}\sqrt{{{x}^{2}}+9}$
We shall use chain rule here taking $u={{x}^{2}}+9$ and $g\left( x \right)=\sqrt{{{x}^{2}}+9}$. We have ;
$\Rightarrow \dfrac{d}{dx}g\left( x \right)=\dfrac{d}{d\left( {{x}^{2}}+9 \right)}\sqrt{{{x}^{2}}+9}\times \dfrac{d}{dx}\left( {{x}^{2}}+9 \right)$
We use the standard differentiation of square function $\dfrac{d}{dt}\sqrt{t}=\dfrac{1}{2\sqrt{t}}$ for $t={{x}^{2}}+9$and sum rule of differentiation to have;

& \Rightarrow \dfrac{d}{dx}g\left( x \right)=\dfrac{1}{2\sqrt{{{x}^{2}}+9}}\left( \dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}9 \right) \\\ & \Rightarrow \dfrac{d}{dx}g\left( x \right)=\dfrac{1}{2\sqrt{{{x}^{2}}+9}}\left( 2x \right) \\\ & \Rightarrow \dfrac{d}{dx}g\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+9}} \\\ \end{aligned}$$ Now we use the quotient rule to have; $$\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\dfrac{g\left( x \right)\dfrac{d}{dx}f\left( x \right)-f\left( x \right)\dfrac{d}{dx}g\left( x \right)}{{{g}^{2}}\left( x \right)} \\\ & \Rightarrow \dfrac{d}{dx}\left( \dfrac{5x}{\sqrt{{{x}^{2}}+9}} \right)=\dfrac{\sqrt{{{x}^{2}}+9}\cdot 5-5x\cdot \dfrac{x}{\sqrt{{{x}^{2}}+9}}}{{{\left( \sqrt{{{x}^{2}}+9} \right)}^{2}}} \\\ \end{aligned}$$ We simplify the terms in numerator and denominator ro have; $$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( \dfrac{5x}{\sqrt{{{x}^{2}}+9}} \right)=\dfrac{\dfrac{\left( {{x}^{2}}+9 \right)\cdot 5-5{{x}^{2}}}{\sqrt{{{x}^{2}}+9}}}{{{x}^{2}}+9} \\\ & \Rightarrow \dfrac{d}{dx}\left( \dfrac{5x}{\sqrt{{{x}^{2}}+9}} \right)=\dfrac{5{{x}^{2}}+45-5{{x}^{2}}}{\left( {{x}^{2}}+9 \right)\sqrt{{{x}^{2}}+9}} \\\ & \Rightarrow \dfrac{d}{dx}\left( \dfrac{5x}{\sqrt{{{x}^{2}}+9}} \right)=\dfrac{45}{{{\left( {{x}^{2}}+9 \right)}^{\dfrac{3}{2}}}} \\\ \end{aligned}$$ **Note:** We note that sum rule of differentiation is given by $\dfrac{d}{dx}\left( f+g \right)=\dfrac{d}{dx}f+\dfrac{d}{dx}g$. We note that here $g\left( x \right)=\sqrt{{{x}^{2}}+9}$ satisfies the condition$g\left( x \right)\ne 0$. Since square is always non-negative we have ${{x}^{2}}\ge 0\Rightarrow {{x}^{2}}+9\ge 9\Rightarrow \sqrt{{{x}^{2}}+9}\ge 3$. So we have always$g\left( x \right)=\sqrt{{{x}^{2}}+9}>0$. Hence the required condition is satisfied. We note that domain of $f\left( x \right)=5x$ is real number set and the domain of $\sqrt{{{x}^{2}}+9}$ is the set $\left[ -3,\infty \right)$ and both $f,g$ are differentiable everywhere in their respective domains. The domain of $\dfrac{f\left( x \right)}{g\left( x \right)}$ is $\left( -3,\infty \right)$.