Derivatives of inverse trigonometric functions
Learn the derivatives of inverse trigonometric functions for Year 12 Specialist Mathematics in Queensland (QCAA). Arcsine, arccosine and arctangent each have a compact algebraic derivative, and the chain rule and product rule extend them to composite and product functions.
You will learn the scaled standard forms \(\dfrac{1}{\sqrt{a^2-x^2}}\) and \(\dfrac{a}{a^2+x^2}\), differentiate composite expressions, and find gradients and tangents to inverse-trig curves — the derivatives that underpin the standard integrals in Unit 4 integration techniques.
Theory
The derivatives of the inverse trigonometric functions arcsine, arccosine and arctangent are three standard results in Year 12 Specialist Mathematics (QCAA, Queensland). Each has a scaled form with a constant \(a\), and the chain rule and product rule extend them to composite and product functions. This page states the formulas and works through differentiating, evaluating and finding tangents.
An inverse trigonometric function undoes a trigonometric function on a restricted domain: arcsine \(\arcsin x\) (also written \(\sin^{-1}x\)), arccosine \(\arccos x\) and arctangent \(\arctan x\). Each returns the angle whose sine, cosine or tangent is \(x\).
Differentiating these gives three results that are purely algebraic — the trigonometric functions disappear. Arcsine and arccosine share the surd \(\sqrt{a^2-x^2}\) and differ only by a sign; arctangent gives a rational function with \(a^2+x^2\).
The constant \(a\) comes from the scaled forms \(\arcsin\dfrac{x}{a}\), \(\arccos\dfrac{x}{a}\) and \(\arctan\dfrac{x}{a}\). Read off \(a\), substitute, and the derivative follows directly.
When the input is a function of \(x\) rather than just \(x\) or \(\dfrac{x}{a}\), apply the chain rule; when the arc-function multiplies another function, apply the product rule. These derivatives are the foundation of the standard integrals met next in this topic.
The three standard derivatives, in scaled form with constant \(a>0\):
With \(a=1\) these give the base results \(\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}\), \(\dfrac{d}{dx}\arccos x=\dfrac{-1}{\sqrt{1-x^2}}\) and \(\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}\). For a function \(u=u(x)\) inside the arc-function, the chain rule multiplies by \(\dfrac{du}{dx}\):
How to differentiate an inverse trig function
- Identify the form: is it \(\arcsin\), \(\arccos\) or \(\arctan\), and is the input \(\dfrac{x}{a}\), a multiple, or a function \(u(x)\)?
- Choose the rule: a scaled input \(\dfrac{x}{a}\) uses the standard formula directly; a function inside uses the chain rule \(\times\dfrac{du}{dx}\); an arc-function multiplying another factor uses the product rule.
- Substitute and simplify: put the input into the formula, then tidy the surd or fraction.
- Evaluate if asked: for a gradient at a point substitute \(x\); for a tangent use the point on the curve and point-gradient form \(y-y_1=m(x-x_1)\).
Read off \(a=5\) and apply \(\dfrac{d}{dx}\arctan\dfrac{x}{a}=\dfrac{a}{a^2+x^2}\):
| \(y\) | \(=\) | \(\arctan\dfrac{x}{5}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{5}{5^2+x^2}\) |
| \(=\) | \(\dfrac{5}{25+x^2}\) |
\(\dfrac{dy}{dx}=\dfrac{5}{25+x^2}\).
Let \(u=4x\), so \(\dfrac{du}{dx}=4\); apply \(\dfrac{d}{dx}\arccos u=\dfrac{-1}{\sqrt{1-u^2}}\cdot\dfrac{du}{dx}\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{-1}{\sqrt{1-u^2}}\times\dfrac{du}{dx}\) |
| \(=\) | \(\dfrac{-1}{\sqrt{1-(4x)^2}}\times 4\) | |
| \(=\) | \(\dfrac{-4}{\sqrt{1-16x^2}}\) |
\(\dfrac{dy}{dx}=\dfrac{-4}{\sqrt{1-16x^2}}\).
Product rule \((uv)'=u'v+uv'\) with \(u=x\) and \(v=\arcsin x\):
| \(u=x,\quad u'\) | \(=\) | \(1\) |
| \(v=\arcsin x,\quad v'\) | \(=\) | \(\dfrac{1}{\sqrt{1-x^2}}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \((1)\arcsin x + x\cdot\dfrac{1}{\sqrt{1-x^2}}\) |
| \(=\) | \(\arcsin x+\dfrac{x}{\sqrt{1-x^2}}\) |
\(\dfrac{dy}{dx}=\arcsin x+\dfrac{x}{\sqrt{1-x^2}}\).
Differentiate, then find the gradient at \(x=\sqrt{3}\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{1}{1+x^2}\) |
| \(m\) | \(=\) | \(\dfrac{1}{1+(\sqrt{3})^2}\) |
| \(=\) | \(\dfrac{1}{4}\) |
Find the point on the curve, using \(\arctan\sqrt{3}=\dfrac{\pi}{3}\):
| \(y_1\) | \(=\) | \(\arctan\sqrt{3}\) |
| \(=\) | \(\dfrac{\pi}{3}\) |
Apply point-gradient form \(y-y_1=m(x-x_1)\):
| \(y-\dfrac{\pi}{3}\) | \(=\) | \(\dfrac{1}{4}\left(x-\sqrt{3}\right)\) |
\(y-\dfrac{\pi}{3}=\dfrac{1}{4}\left(x-\sqrt{3}\right)\).
Common pitfalls
Frequently asked questions
What is the derivative of arcsin x?
It is \(\dfrac{1}{\sqrt{1-x^2}}\). The scaled version is \(\dfrac{d}{dx}\arcsin\dfrac{x}{a}=\dfrac{1}{\sqrt{a^2-x^2}}\).
How is the derivative of arctan x different from arcsin x?
Arctangent gives \(\dfrac{1}{1+x^2}\), a rational function with no surd, whereas arcsine gives \(\dfrac{1}{\sqrt{1-x^2}}\), which has a square-root denominator and a restricted domain.
Why does arccos have a minus sign?
Because \(\arcsin x+\arccos x=\dfrac{\pi}{2}\) is constant, differentiating both sides shows the derivative of \(\arccos x\) is exactly the negative of the derivative of \(\arcsin x\).
Do I use the chain rule for arcsin 2x?
Yes. Set \(u=2x\); then \(\dfrac{d}{dx}\arcsin 2x=\dfrac{1}{\sqrt{1-(2x)^2}}\times 2=\dfrac{2}{\sqrt{1-4x^2}}\).
How do I find the gradient of an inverse trig curve at a point?
Differentiate to get \(\dfrac{dy}{dx}\), then substitute the \(x\)-value. Rationalise any surd so the exact gradient is in simplest form.
Where is the derivative of arcsin(x/a) defined?
Only where \(a^2-x^2>0\), that is on the open interval \(-a