Chapter 3.5, Problem 1E

  $\left(a\right)$

To determine

To calculate: The first derivative of the function $xy+2x+3x^2=4$ by implicit differentiation.

Answer to Problem 1E

The first derivative of the function is $y\text{'}=\frac{-y-6x-2}{x}$ .

Explanation of Solution

Given information:

The function $xy+2x+3x^2=4$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Calculation:

Consider the function $xy+2x+3x^2=4$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(xy+2x+3x^2\right)=\frac{d}{dx}\left(4\right)\\ \frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(3x^2\right)=0\end{array}$

Recall that power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ and chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Apply it. Also observe that y is a function of x,

  $\begin{array}{c}\frac{d}{dx}\left(xy+2x+3x^2\right)=\frac{d}{dx}\left(4\right)\\ \frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(3x^2\right)=0\\ y+xy\text{'}+2+6x=0\end{array}$

Isolate the value of $y\text{'}$ on left hand side and simplify,

  $\begin{array}{c}y+xy\text{'}+2+6x=0\\ xy\text{'}=-y-6x-2\\ y\text{'}=\frac{-y-6x-2}{x}\end{array}$

Thus, the first derivative of the function is $y\text{'}=\frac{-y-6x-2}{x}$ .

  $\left(b\right)$

To determine

To calculate: The first derivative of the function $xy+2x+3x^2=4$ by explicit differentiation.

Answer to Problem 1E

The first derivative of the function is $y\text{'}=\frac{-3x^2-4}{x^2}$ .

Explanation of Solution

Given information:

The function $xy+2x+3x^2=4$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Quotient rule for differentiation is $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ where f and g are functions of x .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Calculation:

Consider the function $xy+2x+3x^2=4$ .

Isolate the value of y on left hand side on the above equation.

Subtract the terms $2x+3x^2$ on both sides of the equation.

  $\begin{array}{c}xy=-2x-3x^2+4\\ y=\frac{-2x-3x^2+4}{x}\end{array}$

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\frac{-2x-3x^2+4}{x}\right)\\ y\text{'}=\frac{d}{dx}\left(\frac{-2x-3x^2+4}{x}\right)\end{array}$

Recall that power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ and chain rule for differentiation is if f is a function of g then $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ also quotient rule for differentiation is $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ where f and g are functions of x .

Apply it. Also observe that y is a function of x,

  $\begin{array}{c}y\text{'}=\frac{x\left(-2x-3x^2+4\right)\text{'}-\left(-2x-3x^2+4\right)\left(x\right)\text{'}}{x^2}\\ y\text{'}=\frac{x\left(-2-6x\right)-\left(-2x-3x^2+4\right)1}{x^2}\\ y\text{'}=\frac{-2x-6x^2+2x+3x^2-4}{x^2}\\ y\text{'}=\frac{-3x^2-4}{x^2}\end{array}$

Thus, the first derivative of the function is $y\text{'}=\frac{-3x^2-4}{x^2}$ .

  $\left(c\right)$

To determine

To verify: The solutions obtained in parts $\left(a\right)\text{ and }\left(b\right)$ are consistent.

Explanation of Solution

Given information:

The derivative of the function $xy+2x+3x^2=4$ by explicit differentiation is $y\text{'}=\frac{-3x^2-4}{x^2}$ and by implicit differentiation is $y\text{'}=\frac{-y-6x-2}{x}$ .

Consider the derivative of the function $xy+2x+3x^2=4$ that is by explicit differentiation it is $y\text{'}=\frac{-3x^2-4}{x^2}$ and by implicit differentiation it is $y\text{'}=\frac{-y-6x-2}{x}$ .

In order to verify that derivative obtained implicitly and explicitly is same substitute the value of y from the function in the value of derivative found by implicit differentiation.

Consider the function $xy+2x+3x^2=4$ .

Isolate the value of y on left hand side on the above equation.

Subtract the terms $2x+3x^2$ on both sides of the equation.

  $\begin{array}{c}xy=-2x-3x^2+4\\ y=\frac{-2x-3x^2+4}{x}\end{array}$

The first derivative of the function by implicit differentiationis $y\text{'}=\frac{-y-6x-2}{x}$ .

Now,

  $\begin{array}{c}y\text{'}=\frac{-\left(\frac{-2x-3x^2+4}{x}\right)-6x-2}{x}\\ =\frac{2x+3x^2-4-6x^2-2x}{x^2}\\ =\frac{-3x^2-4}{x^2}\end{array}$

Since, the derivative found implicitly is equal to derivative found explicitly so both the derivatives are equal.

Hence, the solutions obtained in parts $\left(a\right)\text{ and }\left(b\right)$ are consistent.