7.9 Implicit and parametric differentiation — Practise with help

1. $\dfrac{dy}{dx} = 2x\sin x + x^2\cos x$.
2. $\dfrac{dy}{dx} = 2\cos 2x$.
3. $\dfrac{dy}{dt} = -\sin t$.

If question 1 was hard, revise the product rule in Section 7.7.

Differentiate both sides with respect to $x$. The $y^2$ term becomes $2y\dfrac{dy}{dx}$ by the chain rule.

$2x + 2y\dfrac{dy}{dx} = 0$. Move the $2x$ to the other side and divide by $2y$.

The gradient depends on both coordinates. It is undefined where $y = 0$, at the points $(\pm 4, 0)$, and the tangent is vertical there.

From the pattern of question 1, $\dfrac{dy}{dx} = -\dfrac{x}{y}$. Substitute the coordinates of the given point.

Substitute $x = 6$ and $y = 8$ into $-\dfrac{x}{y}$.

The gradient is negative because the circle is falling at that point in the first quadrant.

The left side is a product, so use the product rule when you differentiate $xy$ with respect to $x$.

Differentiating gives $y + x\dfrac{dy}{dx} = 0$, because the derivative of the constant $12$ is zero.

This matches the derivative of $y = \dfrac{12}{x}$, which is $-\dfrac{12}{x^2} = -\dfrac{y}{x}$.

Differentiate $x^2$ term by term. The $xy$ term needs the product rule, and the constant $6$ differentiates to zero.

$2x + y + x\dfrac{dy}{dx} = 0$. Collect the derivative term on its own.

Differentiate $x$ and $y$ separately with respect to $t$, then divide $\dfrac{dy}{dt}$ by $\dfrac{dx}{dt}$.

$\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 4t^3$. Write the quotient and cancel the common factor.

Cancelling the common factor $2t$ leaves a tidy expression in $t$.

Find the gradient first with $\dfrac{dy}{dx} = -\dfrac{x}{y}$, then use the point and the gradient in $y - y_1 = m(x - x_1)$.

At $(4, 3)$ the gradient is $-\dfrac{4}{3}$. Substitute into the tangent formula.

The tangent is perpendicular to the radius from the origin to $(4, 3)$, which is why its gradient is the negative reciprocal of $\dfrac{3}{4}$.