Question

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

Answer 1

The answer is :

`y^(\prime)=(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 :

`(d)/(dx)[y(\sin x)]=(d)/(dx)(x^3)+(d)/(dx)(\cos y)`

Simplify :

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

That will be :

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