Derivatives of Implicit Trigonometric Functions

This lesson works through four implicit equations that mix $x$ and $y$ inside trigonometric functions. In each one we differentiate both sides with respect to $x$, remembering that every $y$ carries a factor of $\tfrac{dy}{dx}$, then gather the derivative terms and solve.

Example 1: $2x + 3y = \sin y$

Differentiate each side with respect to $x$. The right side uses the chain rule on $\sin y$.

$2 + 3\,\frac{dy}{dx} = \cos y \,\frac{dy}{dx}$

Collect the $\tfrac{dy}{dx}$ terms on one side.

$\frac{dy}{dx}\,(3 - \cos y) = -2$

$\frac{dy}{dx} = \frac{-2}{\,3 - \cos y\,}$

Example 2: $\sin^2 y + \cos(xy) = 1$

Left term. Differentiate $\sin^2 y$ with the chain rule: $2\sin y\cos y\,\tfrac{dy}{dx} = \sin(2y)\,\tfrac{dy}{dx}$.

Right term. Differentiate $\cos(xy)$, using the product rule on $xy$:

$-\sin(xy)\left(x\,\frac{dy}{dx} + y\right)$

Putting it together and setting the sum to $0$:

$\sin(2y)\,\frac{dy}{dx} - \sin(xy)\left(x\,\frac{dy}{dx} + y\right) = 0$

Collect the $\tfrac{dy}{dx}$ terms.

$\frac{dy}{dx}\,\big(\sin 2y - x\sin(xy)\big) = y\sin(xy)$

$\frac{dy}{dx} = \frac{y\sin(xy)}{\,\sin 2y - x\sin(xy)\,}$

Example 3: $\sin^2 x + \cos^2 y = 1$

Differentiate each squared term with the chain rule.

$2\sin x\cos x + 2\cos y(-\sin y)\,\frac{dy}{dx} = 0$

Using the double-angle form $2\sin\theta\cos\theta = \sin 2\theta$:

$\sin 2x - \sin 2y\,\frac{dy}{dx} = 0$

$\frac{dy}{dx} = \frac{\sin 2x}{\sin 2y}$

Example 4: $xy + y^2 = \tan(x+y)$

Differentiate with respect to $x$. The left side uses the product rule on $xy$, and the right side uses the chain rule on $\tan(x+y)$.

$x\,\frac{dy}{dx} + y + 2y\,\frac{dy}{dx} = \sec^2(x+y)\left(1 + \frac{dy}{dx}\right)$

Collect the $\tfrac{dy}{dx}$ terms on one side and the rest on the other.

$\frac{dy}{dx}\,\big(x + 2y - \sec^2(x+y)\big) = \sec^2(x+y) - y$

$\frac{dy}{dx} = \frac{\sec^2(x+y) - y}{\,x + 2y - \sec^2(x+y)\,}$

Key takeaways

• Differentiate both sides with respect to $x$, and every term in $y$ picks up a factor of $\tfrac{dy}{dx}$.
• Use the chain rule on trigonometric functions and the product rule on mixed terms such as $xy$.
• After differentiating, gather all $\tfrac{dy}{dx}$ terms on one side and solve for the derivative.