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

Combining the above yields

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

Solving for $y'$ gives the result

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