Question

Use implicit differentiation to find dy/dx of the equation y(sin x) = x^3+ cos Y

Answer 1

ANSWER :

The answer is :

$y^{\prime}=\frac{3x^2-y\cos x}{\sin x+\sin y}$

EXPLANATION :

From the problem, we have the equation :

$y(\sin x)=x^3+\cos y$

In implicit differentiation, we will differentiate each term :

$\frac{d}{dx}[y(\sin x)]=\frac{d}{dx}(x^3)+\frac{d}{dx}(\cos y)$

Simplify :

$\begin{gathered} \frac{d}{dx}[y(\sin x)]=y(\cos x)+y^{\prime}(\sin x) \\ \\ \frac{d}{dx}(x^3)=3x^2 \\ \\ \frac{d}{dx}(\cos y)=-y^{\prime}\sin y \end{gathered}$

That will be :

$\begin{gathered} \frac{d}{dx}[y(\sin x)]=\frac{d}{dx}(x^3)+\frac{d}{dx}(\cos y) \\ \\ y\cos x+y^{\prime}\sin x=3x^2-y^{\prime}\sin y \end{gathered}$

Express the equation as y' in terms of the other variables.

$\begin{gathered} y\cos x+y^{\prime}\sin x=3x^2-y^{\prime}\sin y \\ y^{\prime}\sin x+y^{\prime}\sin y=3x^2-y\cos x \\ y^{\prime}(\sin x+\sin y)=3x^2-y\cos x \\ \\ y^{\prime}=\frac{3x^2-y\cos x}{\sin x+\sin y} \end{gathered}$