Chapter 3.5, Problem 16E

To determine

To calculate: The derivative of the function $\sin x+\cos y=\sin x\cos y$ by implicit differentiation.

Answer to Problem 16E

The derivative of the function is $y\text{'}=\frac{\cos x\left(\cos y-1\right)}{\sin y\left(\sin x-1\right)}$ .

Explanation of Solution

Given information:

The function $\sin x+\cos y=\sin x\cos y$ .

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 $\sin x+\cos y=\sin x\cos y$ .

Differentiate both sides with respect to x ,

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

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 $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Also for the terms of the above expression, apply the product rule for differentiation.

Recall that 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 .

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

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

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

  $\begin{array}{c}\cos x-\sin yy\text{'}=\cos x\cos y-\sin x\sin yy\text{'}\\ \left(\sin x\sin y-\sin y\right)y\text{'}=\cos x\cos y-\cos x\\ y\text{'}=\frac{\cos x\cos y-\cos x}{\sin x\sin y-\sin y}\\ y\text{'}=\frac{\cos x\left(\cos y-1\right)}{\sin y\left(\sin x-1\right)}\end{array}$

Thus, the derivative of the function is $y\text{'}=\frac{\cos x\left(\cos y-1\right)}{\sin y\left(\sin x-1\right)}$ .