Chapter 3.5, Problem 2E

  $\left(a\right)$

To determine

To calculate: The first derivative of the function $\cos x+\sqrt{y}=5$ by implicit differentiation.

Answer to Problem 2E

The first derivative of the function is $y\text{'}=2\sin x\sqrt{y}$ .

Explanation of Solution

Given information:

The function $\cos x+\sqrt{y}=5$ .

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 $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Calculation:

Consider the function $\cos x+\sqrt{y}=5$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(\cos x+\sqrt{y}\right)=\frac{d}{dx}\left(5\right)\\ \frac{d}{dx}\left(\cos x\right)+\frac{d}{dx}\left(\sqrt{y}\right)=0\end{array}$

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,

  $\begin{array}{c}\frac{d}{dx}\left(\cos x+\sqrt{y}\right)=\frac{d}{dx}\left(5\right)\\ \frac{d}{dx}\left(\cos x\right)+\frac{d}{dx}\left(\sqrt{y}\right)=0\\ -\sin x+\frac{1}{2\sqrt{y}}y\text{'}=0\end{array}$

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

  $\begin{array}{c}-\sin x+\frac{1}{2\sqrt{y}}y\text{'}=0\\ \frac{1}{2\sqrt{y}}y\text{'}=\sin x\\ y\text{'}=2\sin x\sqrt{y}\end{array}$

Thus, the first derivative of the function is $y\text{'}=2\sin x\sqrt{y}$ .

  $\left(b\right)$

To determine

To calculate: The first derivative of the function $\cos x+\sqrt{y}=5$ by explicit differentiation.

Answer to Problem 2E

The first derivative of the function is $y\text{'}=2\sin x\left(5-\cos x\right)$ .

Explanation of Solution

Given information:

The function $\cos x+\sqrt{y}=5$ .

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 $\cos x+\sqrt{y}=5$ .

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

Subtract the terms $\cos x$ on both sides of the equation.

  $\sqrt{y}=5-\cos x$

Square both the sides of the above expression,

  $\begin{array}{c}y={\left(5-\cos x\right)}^2\\ =25+{\cos }^{2}x-10\cos x\end{array}$

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(25+{\cos }^{2}x-10\cos x\right)\\ y\text{'}=\frac{d}{dx}\left(25\right)+\frac{d}{dx}\left({\cos }^{2}x\right)-\frac{d}{dx}\left(10\cos x\right)\end{array}$

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 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,

  $\begin{array}{c}y\text{'}=0-2\cos x\sin x+10\sin x\\ y\text{'}=2\sin x\left(5-\cos x\right)\end{array}$

Thus, the first derivative of the function is $y\text{'}=2\sin x\left(5-\cos x\right)$ .

  $\left(c\right)$

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 $\cos x+\sqrt{y}=5$ by explicit differentiation is $y\text{'}=2\sin x\left(5-\cos x\right)$ and by implicit differentiation is $y\text{'}=2\sin x\sqrt{y}$ .

Consider the derivative of the function $\cos x+\sqrt{y}=5$ that is by explicit differentiation it is $y\text{'}=2\sin x\left(5-\cos x\right)$ and by implicit differentiation it is $y\text{'}=2\sin x\sqrt{y}$ .

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 $\cos x+\sqrt{y}=5$ .

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

Subtract the terms $\cos x$ on both sides of the equation.

  $\sqrt{y}=5-\cos x$

The first derivative of the function by implicit differentiation is $y\text{'}=2\sin x\sqrt{y}$ .

Now,

  $\begin{array}{c}y\text{'}=2\sin x\sqrt{y}\\ y\text{'}=2\sin x\left(5-\cos x\right)\end{array}$

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.