Implicit, logarithmic and parametric differentiation

What you will be able to do

• differentiate an equation in $x$ and $y$ without solving for $y$;
• use logarithmic differentiation for $x^x$ and for long products and quotients;
• find the slope of a curve given by a parameter.

Implicit differentiation

Some curves are given by an equation in $x$ and $y$, such as the circle $x^2 + y^2 = 25$. Solving for $y$ gives two separate branches. Implicit differentiation avoids this. Differentiate both sides with respect to $x$, and remember that $y$ is a function of $x$. By the chain rule,

For the circle, $2x + 2y\dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -\dfrac{x}{y}$. At the point $(3, 4)$ the slope is $-\dfrac{3}{4}$.

Logarithmic differentiation

Two cases are awkward: a power with $x$ in the exponent, such as $x^x$, and a long product or quotient. In both, take the natural logarithm first.

1. Write $\ln y$ in terms of $x$. Use $\ln(uv) = \ln u + \ln v$ and $\ln(u^v) = v\ln u$.
2. Differentiate both sides. The left side gives $\dfrac{1}{y}\dfrac{dy}{dx}$.
3. Multiply by $y$ and replace $y$ by its formula.

For $y = x^x$: $\ln y = x\ln x$, so $\dfrac{y'}{y} = \ln x + 1$, and $y' = x^x(1 + \ln x)$.

Parametric differentiation

A curve can be given by a parameter $t$: $x = f(t)$ and $y = g(t)$. The chain rule gives $\dfrac{dy}{dt} = \dfrac{dy}{dx}\times\dfrac{dx}{dt}$. So

For the circle $x = a\cos t$, $y = a\sin t$ this gives $\dfrac{dy}{dx} = \dfrac{a\cos t}{-a\sin t} = -\cot t$.

Worked examples

Implicit. Find $\dfrac{dy}{dx}$ at $(1, 2)$ on the curve $x^2 + xy + y^2 = 7$.

1. Check the point. $1 + 2 + 4 = 7$.
2. Differentiate. $2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$.
3. Solve. $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$. At $(1, 2)$ this is $-\dfrac{4}{5}$.

Parametric. Find the slope of $x = 3\cos t$, $y = 2\sin t$ at $t = \dfrac{\pi}{3}$.

1. The rates. $\dfrac{dx}{dt} = -3\sin t$ and $\dfrac{dy}{dt} = 2\cos t$.
2. The slope. $\dfrac{dy}{dx} = -\dfrac{2}{3}\cot t$.
3. Put in $t$. $\cot\dfrac{\pi}{3} = \dfrac{1}{\sqrt{3}}$, so the slope is $-\dfrac{2}{3\sqrt{3}} \approx -0.385$.

Practise with new numbers

Try it in Python

The program checks all three methods. It compares each exact slope with a numerical slope: the circle at $(3, 4)$, $x^x$ at $x = 2$, and the ellipse of the worked example.

Common mistakes

• Forgetting $\dfrac{dy}{dx}$. The derivative of $y^2$ with respect to $x$ is $2y\dfrac{dy}{dx}$, not $2y$.
• The product rule for $xy$. Both factors depend on $x$, so $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$.
• The power rule on $x^x$. The exponent also moves. Take logarithms.
• Dividing the wrong way. In parametric form, $\dfrac{dy}{dx}$ is $\dfrac{dy}{dt}$ divided by $\dfrac{dx}{dt}$, not the other way round.

Remember these

• Implicit: differentiate both sides, treat $y$ as a function of $x$, then solve for $\dfrac{dy}{dx}$.
• Logarithmic: take $\ln$ of both sides, differentiate, then multiply by $y$. For $x^x$ the result is $x^x(1 + \ln x)$.
• Parametric: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\dfrac{dy}{dx}$ at $(3, 4)$ on the curve $xy = 12$.

2. Differentiate $y = x^{x^2}$ for $x > 0$.

3. For $x = t^2$ and $y = 2t$, find $\dfrac{dy}{dx}$ in terms of $t$.

4. Use logarithmic differentiation to find $y'(0)$ for $y = (x + 1)^2(x + 2)^3$.

5. Find $\dfrac{dy}{dx}$ if $\sin y = x$, for $y$ in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

References

• OpenStax, Calculus Volume 1, Chapter 3, sections on implicit differentiation and on derivatives of exponential and logarithmic functions.
• NCERT, Mathematics Part I for Class XII, Chapter 5, sections on logarithmic differentiation and on derivatives of functions in parametric forms.