Science:Math Exam Resources/Courses/MATH110/April 2019/Question 04 (a)/Solution 1

The procedure is to take the derivative with respect to $x$ of both sides of the equation and then solve for $\frac{dy}{dx}$.

$\frac{d}{dx}(4x^3+x(y-3)+(y-3{)}^3)=\frac{d}{dx}(32).$

The right hand side of the equation above is zero because 32 is a constant.

To compute the left hand side, treat y as a function of x and use the chain rule every time you need to take its derivative:

$\frac{d}{dx}(4x^3+x(y-3)+(y-3{)}^3)=\frac{d}{dx}(4x^3)+\frac{d}{dx}(x(y-3))+\frac{d}{dx}((y-3{)}^3).$

• The derivative $(4x^3{)}^{\prime }=12x^2$ follows from the power rule.
• To compute the second term, use the product rule: $\frac{d}{dx}(x(y-3))=\frac{d}{dx}(x)\cdot (y-3)+x\cdot \frac{d}{dx}(y-3)=1\cdot (y-3)+x(y^{\prime }-0)$.
• The third term is computed using the chain rule: $((y-3{)}^3{)}^{\prime }=3(y-3{)}^2{y}^{\prime }$.

Combining the above yields

$12x^2+y-3+x{y}^{\prime }+3(y-3{)}^2{y}^{\prime }=0.$

Solving for $y^{\prime }$ gives the result

$y^{\prime }=\frac{3-y-12x^2}{x+3(y-3{)}^2}.$