Q. 6 - Calculus

Question

Suppose $u\left(x\right)=\sqrt{3x^2+1}$ and $f\left(u\right)=\frac{u^2+3u^5}{1-u}$. Use the chain rule to find $\frac{d }{dx} \left(f\right(u\left(x\right)\left)\right)$ without first finding the formula for $f\left(u\right(x\left)\right)$. 

Step-by-Step Solution

Answer

The derivative of the function is:$\frac{3x\left(45x^2\sqrt{3x^2+1}+14\sqrt{3x^2+1}-36x^4-72x^2-10\right)}{{\left(1-\sqrt{3x^2+1}\right)}^2}$

Step 1. Given information:

The functions are:

$u\left(x\right)=\sqrt{3x^2+1}$

$f=\frac{u^2+3u^5}{1-u}$

The composite function is:$f\left(u\right(x\left)\right)$

Step 2. Find derivative of u ( x ) using chain rule.

Since, $$\begin{gathered} u\left(x\right)=\sqrt{3x^2+1} \\ u\left(x\right)={(3x^2+1)}^{\frac{1}{2}} \end{gathered}$$

Let $v=3x^2+1$

Then $u={\left(v\right)}^{\frac{1}{2}}$

Derivative of $u\left(v\right)$ with respect to $v$:

$$\begin{gathered} \frac{du}{dv}=\frac{1}{2}{\left(v\right)}^{\frac{1}{2}-1} \\ =\frac{1}{2}{\left(v\right)}^{-\frac{1}{2}} \\ =\frac{1}{2\sqrt{v}} \\ =\frac{1}{2\sqrt{3x^2+1}} \end{gathered}$$

Derivative of $v\left(x\right)$ with respect to $x$:

$$\begin{gathered} \frac{dv}{dx}=\frac{d}{dx}(3x^2+1) \\ =3.2x+0 \\ =6x \end{gathered}$$

$$\begin{gathered} \frac{du}{dx}=\frac{du}{dv}\times \frac{dv}{dx} \\ =\frac{1}{2\sqrt{3x^2+1}}\times 6x \\ =\frac{3x}{\sqrt{3x^2+1}} \end{gathered}$$

Step 3. Find the derivative of f ( u ) :

$f\left(u\right)=\frac{u^2+3u^5}{1-u}$

Differentiate $$\begin{gathered} \frac{df}{du}=\frac{(1-u){\displaystyle \frac{d}{du}}(u^2+3u^5)-(u^2+3u^5){\displaystyle \frac{d}{du}}(1-u)}{{(1-u)}^2} \\ =\frac{(1-u)(2u+15u^4)-(u^2+3u^5)(-1)}{{(1-u)}^2} \\ =\frac{2u+15u^4-2u^2-15u^5+u^2+3u^5}{{(1-u)}^2} \\ =\frac{2u+15u^4-u^2-12u^5}{{(1-u)}^2} \end{gathered}$$

Step 4. Find derivative of f ( u ( x ) ) :

By using the chain rule, the derivative of the function is:

$\frac{d}{dx}\left(f\right(u\left(x\right)\left)\right)=\frac{df}{du}\times \frac{du}{dx}$

Substitute the values from step 3 and 4.

$\frac{d}{dx}\left(f\left(u\right(x\left)\right)\right)=\frac{2u+15u^4-u^2-12u^5}{{(1-u)}^2}\times \frac{3x}{\sqrt{3x^2+1}}$

Substitute value of $u\left(x\right)=\sqrt{3x^2+1}$ and simplify:

$$\begin{gathered} \frac{d}{dx}\left(f\right(u\left(x\right)\left)\right)=\frac{2\sqrt{3x^2+1}+15{\left(\sqrt{3x^2+1}\right)}^4-{\left(\sqrt{3x^2+1}\right)}^2-12{\left(\sqrt{3x^2+1}\right)}^5}{{\left(1-\sqrt{3x^2+1}\right)}^2}\times \frac{3x}{\sqrt{3x^2+1}} \\ =\frac{\sqrt{3x^2+1}\left(2+15{\left(\sqrt{3x^2+1}\right)}^3-\sqrt{3x^2+1}-12{\left(\sqrt{3x^2+1}\right)}^4\right)}{{\left(1-\sqrt{3x^2+1}\right)}^2}\times \frac{3x}{\sqrt{3x^2+1}} \\ =\frac{3x\left(2+15{\left(\sqrt{3x^2+1}\right)}^3-\sqrt{3x^2+1}-12{\left(\sqrt{3x^2+1}\right)}^4\right)}{{\left(1-\sqrt{3x^2+1}\right)}^2} \\ =\frac{3x\left(2+15\left(3x^2+1\right)\sqrt{3x^2+1}-\sqrt{3x^2+1}-12{\left(3x^2+1\right)}^2\right)}{{\left(1-\sqrt{3x^2+1}\right)}^2} \\ =\frac{3x\left(2+45x^2\sqrt{3x^2+1}+15\sqrt{3x^2+1}-\sqrt{3x^2+1}-12\left(3x^4+1+6x^2\right)\right)}{{\left(1-\sqrt{3x^2+1}\right)}^2} \\ =\frac{3x\left(2+45x^2\sqrt{3x^2+1}+14\sqrt{3x^2+1}-36x^4-12-72x^2\right)}{{\left(1-\sqrt{3x^2+1}\right)}^2} \\ =\frac{3x\left(45x^2\sqrt{3x^2+1}+14\sqrt{3x^2+1}-36x^4-72x^2-10\right)}{{\left(1-\sqrt{3x^2+1}\right)}^2} \end{gathered}$$