Chapter 3.5, Problem 15E

To determine

To calculate: The derivative of the function $e^y\cos x=1+\sin \left(xy\right)$ by implicit differentiation.

Answer to Problem 15E

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

Explanation of Solution

Given information:

The function $e^y\cos x=1+\sin \left(xy\right)$ .

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 $e^y\cos x=1+\sin \left(xy\right)$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(e^y\cos x\right)=\frac{d}{dx}\left(1+\sin \left(xy\right)\right)$

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,

  $\begin{array}{c}\frac{d}{dx}\left(e^y\cos x\right)=\frac{d}{dx}\left(1+\sin \left(xy\right)\right)\\ e^y\cos xy\text{'}-e^y\sin x=\cos \left(xy\right)\left[y+xy\text{'}\right]\\ e^y\cos xy\text{'}-e^y\sin x=y\cos \left(xy\right)+x\cos \left(xy\right)y\text{'}\end{array}$

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

  $\begin{array}{c}{e}^y\cos xy\text{'}-e^y\sin x=y\cos \left(xy\right)+x\cos \left(xy\right)y\text{'}\\ \left(e^y\cos x-x\cos \left(xy\right)\right)y\text{'}=y\cos \left(xy\right)+e^y\sin x\\ y\text{'}=\frac{y\cos \left(xy\right)+e^y\sin x}{e^y\cos x-x\cos \left(xy\right)}\end{array}$

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