### Chain Rule & Integration by Substitution

Many ways problem

## Things you might have done

Find the gradient of the curve $x^2+y^2=4$ at the following three points.

There are different ways we could think about finding the gradient of a circle at a point.

What is the derivative at a general point on the curve?

Again, we can tackle this in a number of different ways.

You might notice that the gradient function is in terms of both $x$ and $y$. This makes sense because for each value of $x$, there may be two points on the curve, each of which has a different $y$ and different gradient.

You might check this result by substituting the coordinates given earlier to make sure we get the same gradients.

What happens when $y = 0$? What does this mean geometrically on the circle?

In this case, we had a choice of three methods, but for many other curves the first two approaches do not work and implicit differentiation becomes a valuable tool.

What is the derivative at a general point on this curve?

Can you use the same methods as above?

The first approach above relied on the geometry of the circle. It used the fact that a tangent to a circle is at right-angles to the radius. Is this true for an ellipse? Imagine your ellipse as a circle that has been stretched in one direction. What happens to the angle between the tangent and the radius?

Here we will differentiate implicitly.

\begin{align*} \dfrac{d}{dx}\left(9x^2 + 16y^2 \right) &= \dfrac{d}{dx}(144) \\ 18x + 32y\frac{dy}{dx} &= 0 \end{align*}

We can rearrange to find the gradient.

\begin{align*} \dfrac{dy}{dx} &= -\dfrac{18x}{32y} \\ &= -\dfrac{9x}{16y} \end{align*}