Q. 7 - Calculus

Question

Differentiate $f\left(x\right)={\left(3x+\sqrt{x}\right)}^2$in three ways. When you have completed all three parts, show that your three answers are the same:

(a) with the chain rule 

(b) with the product rule but not the chain rule

 (c) without the chain or product rules.

Step-by-Step Solution

Answer

The derivative of $f\left(x\right)$ using each rule is:$f\text{'}\left(x\right)=18x+9\sqrt{x}+1$

Step 1. Given information:

The function is:

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

Part (a). Step 1. To find derivative using chain rule:

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

Let $u=3x+\sqrt{x}$

then $f\left(u\right)=u^2$

Derivative of $u$ with respect to $x$:

$$\begin{gathered} \frac{du}{dx}=\frac{d}{dx}(3x+\sqrt{x}) \\ =3+\frac{1}{2\sqrt{x}} \end{gathered}$$

Derivative of $f$ with respect to $u$:

$$\begin{gathered} \frac{df}{du}=\frac{d}{du}\left(u^2\right) \\ =2u \\ =2(3x+\sqrt{x}) \end{gathered}$$

Derivative of $f$ with respect to $x$:

$$\begin{gathered} \frac{df}{dx}=\frac{df}{du}\times \frac{du}{dx} \\ =2\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right) \\ =2\left(9x+\frac{3}{2}\sqrt{x}+3\sqrt{x}+\frac{1}{2}\right) \\ =2\left(9x+\frac{9}{2}\sqrt{x}+\frac{1}{2}\right) \\ =18x+9\sqrt{x}+1 \end{gathered}$$

Part (b). Step 1. To find derivative using product rule:

$$\begin{gathered} f\left(x\right)={\left(3x+\sqrt{x}\right)}^2 \\ =\left(3x+\sqrt{x}\right).\left(3x+\sqrt{x}\right) \end{gathered}$$

$$\begin{gathered} U=\left(3x+\sqrt{x}\right) \\ V=\left(3x+\sqrt{x}\right) \end{gathered}$$

$$\begin{gathered} \frac{dU}{dx}=\frac{d}{dx}(3x+\sqrt{x}) \\ =3+\frac{1}{2\sqrt{x}} \\ =\frac{dV}{dx} \end{gathered}$$

Now derivative of a function using the product rule is:

$$\begin{gathered} \frac{df}{dx}=U\frac{dV}{dx}+V\frac{dU}{dx} \\ =\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right)+\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right) \\ =2\left(3x+\sqrt{x}\right)\left(3+\frac{1}{2\sqrt{x}}\right) \\ =2\left(9x+\frac{3}{2}\sqrt{x}+3\sqrt{x}+\frac{1}{2}\right) \\ =2\left(9x+\frac{9}{2}\sqrt{x}+\frac{1}{2}\right) \\ =18x+9\sqrt{x}+1 \end{gathered}$$

Part (c). Step 1. To find derivative without using chain rule or product rule:

$$\begin{gathered} f\left(x\right)={\left(3x+\sqrt{x}\right)}^2 \\ ={\left(3x\right)}^2+{\left(\sqrt{x}\right)}^2+2\left(3x\right)\left(\sqrt{x}\right) \\ =9x^2+x+6x^{\frac{3}{2}} \end{gathered}$$

Now differentiate this with respect to $x$:

$$\begin{gathered} \frac{df}{dx}=\frac{d}{dx}\left(9x^2+x+6x^{\frac{3}{2}}\right) \\ =9.2x+1+6.\frac{3}{2}{x}^{\frac{1}{2}} \\ =18x+1+9\sqrt{x} \end{gathered}$$