IB Maths AA 5.8 Notes

Implicit differentiation

Implicit differentiation is used when an equation is not written with $y$ on its own. Instead of first rearranging into the form $y=f(x)$, we differentiate both sides with respect to $x$ and treat $y$ as a function of $x$.

For example, in an equation such as $x^2+y^2=25$, the variable $y$ depends on $x$ even though it is not isolated.

Some curves are difficult or impossible to write as a single function of $x$. A circle is a simple example. The equation $x^2+y^2=25$ gives both an upper semicircle and a lower semicircle, so it is often easier to differentiate implicitly.

While it sounds difficult, the key idea is that you apply the chain rule. Thus, when differentiating a term involving $y$, multiply by $\frac{dy}{dx}$ because of the chain rule. For example:

• $\frac{d}{dx}(y)=\frac{dy}{dx}$
• $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$
• $\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}$

Differentiate $x^2+y^2=25$ implicitly.

Differentiate both sides with respect to $x$: $2x+2y\frac{dy}{dx}=0$.

Solve for $\frac{dy}{dx}$: $2y\frac{dy}{dx}=-2x$.

So $\frac{dy}{dx}=-\frac{x}{y}$.

Differentiate $x^3+y^3=6xy$ implicitly.

Differentiate both sides: $3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}$.

Collect the $\frac{dy}{dx}$ terms: $3y^2\frac{dy}{dx}-6x\frac{dy}{dx}=6y-3x^2$.

Factorise: $\frac{dy}{dx}(3y^2-6x)=6y-3x^2$.

So $\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}$.

Once $\frac{dy}{dx}$ has been found, substitute the coordinates of a point to find the gradient there.

Find the gradient of $x^2+y^2=25$ at $(3,4)$.

From earlier, $\frac{dy}{dx}=-\frac{x}{y}$.

Substitute $(3,4)$ to get $\frac{dy}{dx}=-\frac{3}{4}$.