How do you differentiate a product like x sin x or e^x sin x?

Use the product rule: the derivative of u times v is u prime v plus u v prime. For x sin x this gives sin x plus x cos x. For e to the x times sin x it gives e to the x times sin x plus e to the x times cos x, which factorises to e to the x times (sin x plus cos x).

How do you find the gradient of a trig curve at an exact angle?

The gradient is the derivative evaluated at that angle. For y = sin x the derivative is cos x, so at x = pi over 3 the gradient is cos(pi over 3) = one half. That value is the slope of the tangent line at the point.

How do you differentiate a product like x sin x or e^x sin x?

Use the product rule: the derivative of u times v is u prime v plus u v prime. For x sin x this gives sin x plus x cos x. For e to the x times sin x it gives e to the x times sin x plus e to the x times cos x, which factorises to e to the x times (sin x plus cos x).

How do you find the gradient of a trig curve at an exact angle?

The gradient is the derivative evaluated at that angle. For y = sin x the derivative is cos x, so at x = pi over 3 the gradient is cos(pi over 3) = one half. That value is the slope of the tangent line at the point.

Year 12 Maths - Methods (Unit 3 & Unit 4) Differentiation

Differentiation of trigonometric functions

Q: What is the derivative of sin x and cos x?

In radians, the derivative of sin x is cos x, and the derivative of cos x is negative sin x. The minus sign belongs to the derivative of cosine. The derivative of tan x is sec squared x, which equals 1 divided by cos squared x.

Q: How do you differentiate sin(f(x)) or cos(f(x))?

Use the chain rule: the derivative of sin(f(x)) is f prime of x times cos(f(x)), and the derivative of cos(f(x)) is negative f prime of x times sin(f(x)). Differentiate the outer trig function, then multiply by the derivative of the inside. For example the derivative of cos(3x) is negative 3 sin(3x).

Q: Why must the angle be in radians?

The rules d/dx sin x = cos x and d/dx cos x = -sin x are only true when x is measured in radians. To differentiate sin of x degrees, first convert to radians: sin x degrees equals sin of (pi x over 180), so its derivative is (pi over 180) cos x degrees.

Q: How do you find stationary points of a trig function?

Differentiate and set the derivative equal to zero, then solve over the given domain. For y = sin x + cos x the derivative is cos x minus sin x; setting it to zero gives tan x = 1, so x = pi over 4 and 5 pi over 4 on the interval 0 to 2 pi, giving a maximum of root 2 and a minimum of negative root 2.

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Theory

Differentiating trigonometric functions (in radians) uses the standard rules \(\dfrac{d}{dx}\sin x=\cos x\), \(\dfrac{d}{dx}\cos x=-\sin x\) and \(\dfrac{d}{dx}\tan x=\sec^{2}x\). For a composite such as \(\sin(3x)\) use the chain rule, and combine with the product and quotient rules when a trig function multiplies a polynomial or \(e^{x}\).

The derivative of a trigonometric function gives the gradient of its curve. In radians, the three standard results are \(\dfrac{d}{dx}\sin x=\cos x\), \(\dfrac{d}{dx}\cos x=-\sin x\) and \(\dfrac{d}{dx}\tan x=\sec^{2}x=\dfrac{1}{\cos^{2}x}\). The minus sign belongs to the derivative of cosine; the \(\tan\) result follows from the quotient rule on \(\tan x=\dfrac{\sin x}{\cos x}\).

For a composite function, the chain rule applies: \(\dfrac{d}{dx}\sin\!\big(f(x)\big)=f'(x)\cos\!\big(f(x)\big)\) and \(\dfrac{d}{dx}\cos\!\big(f(x)\big)=-f'(x)\sin\!\big(f(x)\big)\). You differentiate the outer trig function and multiply by the derivative of the inside — never dropping that inner factor.

When a trig function is multiplied by (or divided by) another function — a polynomial or \(e^{x}\) — use the product rule \((uv)'=u'v+uv'\) or the quotient rule. Evaluating a derivative at an exact-value angle (\(\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3}\)) gives an exact gradient, used to find tangents and stationary points.

Key idea. \(\dfrac{d}{dx}\sin x=\cos x\), \(\dfrac{d}{dx}\cos x=-\sin x\), \(\dfrac{d}{dx}\tan x=\sec^{2}x\); for \(\sin\!\big(f(x)\big)\) or \(\cos\!\big(f(x)\big)\) multiply by \(f'(x)\).

Differentiating \(y=\sin x\) gives its gradient function \(y'=\cos x\)

Gradient of \(y=\sin x\) at \(x=\tfrac{\pi}{3}\) is \(\cos\tfrac{\pi}{3}=\tfrac12\)

The three standard derivatives (radians):

\[\frac{d}{dx}\sin x=\cos x \qquad \frac{d}{dx}\cos x=-\sin x \qquad \frac{d}{dx}\tan x=\sec^{2}x=\frac{1}{\cos^{2}x}\]

\frac{d}{d x} \sin x = \cos x

The chain rule for a composite trig function:

\[\frac{d}{dx}\sin\!\big(f(x)\big)=f'(x)\cos\!\big(f(x)\big) \qquad \frac{d}{dx}\cos\!\big(f(x)\big)=-f'(x)\sin\!\big(f(x)\big)\]

\frac{d}{d x} \sin (f (x)) = f^{'} (x) \cos (f (x))

The product rule (a trig function times another function):

\[(uv)'=u'v+uv' \qquad \frac{d}{dx}\big(x\sin x\big)=\sin x+x\cos x\]

(u v)^{'} = u^{'} v + u v^{'}

Degrees. The rules hold only in radians. For \(\sin x^{\circ}\), convert first: \(\sin x^{\circ}=\sin\dfrac{\pi x}{180}\), so \(\dfrac{d}{dx}\sin x^{\circ}=\dfrac{\pi}{180}\cos x^{\circ}\).

How to differentiate a trigonometric function

Check it is in radians. If the angle is in degrees, rewrite it in radians first (\(x^{\circ}=\tfrac{\pi x}{180}\)) before differentiating.
Apply the standard rule. \(\dfrac{d}{dx}\sin x=\cos x\), \(\dfrac{d}{dx}\cos x=-\sin x\), \(\dfrac{d}{dx}\tan x=\sec^{2}x\). Keep the minus sign on the cosine.
Composite? Use the chain rule. For \(\sin\!\big(f(x)\big)\) or \(\cos\!\big(f(x)\big)\), multiply by \(f'(x)\); e.g. \(\dfrac{d}{dx}\cos(3x)=-3\sin(3x)\).
Product or quotient? Combine the rules. When trig multiplies a polynomial or \(e^{x}\), use \((uv)'=u'v+uv'\) (or the quotient rule), then express in simplest and factorised form.
Need a gradient, tangent or stationary point? Evaluate the derivative at the point for the gradient; set \(y'=0\) and solve to locate stationary points.

Simplest form. The syllabus asks for derivatives in simplest and factorised form — e.g. write \(e^{x}\sin x+e^{x}\cos x\) as \(e^{x}(\sin x+\cos x)\).

Example 1 — standard and chain rule

Differentiate (a) \(y=4\sin x-\cos x\) and (b) \(y=\sin(2x-1)\).

Solution

(a) \(y=4\sin x-\cos x\)

Differentiate term by term — \(\dfrac{d}{dx}\sin x=\cos x\), \(\dfrac{d}{dx}\cos x=-\sin x\):

\(y'\)	\(=\)	\(4\cos x-(-\sin x)\)
	\(=\)	\(4\cos x+\sin x\)

\(\therefore\) \(y'=4\cos x+\sin x\)

(b) \(y=\sin(2x-1)\)

Chain rule — the inner \(2x-1\) has derivative \(2\):

\(y'\)	\(=\)	\(\cos(2x-1)\times 2\)
	\(=\)	\(2\cos(2x-1)\)

\(\therefore\) \(y'=2\cos(2x-1)\)

y^{'} = 2 \cos (2 x - 1)

Example 2 — product rule (with \(e^{x}\))

Differentiate \(y=e^{x}\sin x\) and express in factorised form.

Solution

Set up the product rule — \(u=e^{x}\), \(v=\sin x\):

\(u'\)	\(=\)	\(e^{x}\)
\(v'\)	\(=\)	\(\cos x\)

Apply \((uv)'=u'v+uv'\):

\(y'\)

\(=\)

\(e^{x}\sin x+e^{x}\cos x\)

Factorise — take out \(e^{x}\):

\(y'\)

\(=\)

\(e^{x}(\sin x+\cos x)\)

\(\therefore\) \(y'=e^{x}(\sin x+\cos x)\)

y^{'} = e^{x} (\sin x + \cos x)

Example 3 — tangent at an exact angle

Find the equation of the tangent to \(y=\cos x\) at \(x=\dfrac{\pi}{2}\).

Solution

Point — substitute \(x=\dfrac{\pi}{2}\) into the curve:

\(y\!\left(\dfrac{\pi}{2}\right)\)	\(=\)	\(\cos\dfrac{\pi}{2}\)
	\(=\)	\(0\)

Point of contact: \(\left(\dfrac{\pi}{2},\ 0\right)\).

Gradient — differentiate, then substitute:

\(y'\)	\(=\)	\(-\sin x\)
\(m=y'\!\left(\dfrac{\pi}{2}\right)\)	\(=\)	\(-\sin\dfrac{\pi}{2}=-1\)

Line — use \(y-y_{1}=m(x-x_{1})\):

\(y-0\)	\(=\)	\(-1\left(x-\dfrac{\pi}{2}\right)\)
\(y\)	\(=\)	\(-x+\dfrac{\pi}{2}\)

\(\therefore\) Tangent: \(y=-x+\dfrac{\pi}{2}\)

y = - x + \frac{π}{2}

Example 4 — stationary points

Find the stationary points of \(y=\sin x+\cos x\) on \([0,2\pi]\).

Solution

Set the derivative to zero — stationary points have \(y'=0\):

\(y'\)	\(=\)	\(\cos x-\sin x\)
\(\cos x-\sin x\)	\(=\)	\(0\)

Solve for \(x\) on \([0,2\pi]\):

\(\cos x\)	\(=\)	\(\sin x\)
\(\tan x\)	\(=\)	\(1\)
\(x\)	\(=\)	\(\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\)

At \(x=\dfrac{\pi}{4}\)

\(y\)	\(=\)	\(\sin\dfrac{\pi}{4}+\cos\dfrac{\pi}{4}\)
	\(=\)	\(\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}=\sqrt2\)

Maximum \(\left(\dfrac{\pi}{4},\ \sqrt2\right)\).

At \(x=\dfrac{5\pi}{4}\)

\(y\)	\(=\)	\(\sin\dfrac{5\pi}{4}+\cos\dfrac{5\pi}{4}\)
	\(=\)	\(-\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt2}=-\sqrt2\)

Minimum \(\left(\dfrac{5\pi}{4},\ -\sqrt2\right)\).

\(\therefore\) maximum \(\sqrt2\) at \(x=\dfrac{\pi}{4}\); minimum \(-\sqrt2\) at \(x=\dfrac{5\pi}{4}\)

x = \frac{π}{4}, \frac{5 π}{4}

Common pitfalls

Keep the minus sign on \(\cos\). \(\dfrac{d}{dx}\cos x=-\sin x\). Writing \(+\sin x\) is the most frequent error.

Do not drop the inner factor. \(\dfrac{d}{dx}\cos(3x)=-3\sin(3x)\), not \(-\sin(3x)\). The chain rule multiplies by \(f'(x)\).

Radians, not degrees. The rules assume radians. To differentiate \(\sin x^{\circ}\), rewrite as \(\sin\dfrac{\pi x}{180}\) first, giving \(\dfrac{\pi}{180}\cos x^{\circ}\).

Frequently asked questions

What is the derivative of sin x and cos x?

In radians, \(\dfrac{d}{dx}\sin x=\cos x\) and \(\dfrac{d}{dx}\cos x=-\sin x\). Also \(\dfrac{d}{dx}\tan x=\sec^{2}x=\dfrac{1}{\cos^{2}x}\).

How do you differentiate sin(f(x)) or cos(f(x))?

Chain rule: \(\dfrac{d}{dx}\sin\!\big(f(x)\big)=f'(x)\cos\!\big(f(x)\big)\) and \(\dfrac{d}{dx}\cos\!\big(f(x)\big)=-f'(x)\sin\!\big(f(x)\big)\). E.g. \(\dfrac{d}{dx}\cos(3x)=-3\sin(3x)\).

Why must the angle be in radians?

The rules only hold in radians. For \(\sin x^{\circ}\), rewrite as \(\sin\dfrac{\pi x}{180}\); then \(\dfrac{d}{dx}\sin x^{\circ}=\dfrac{\pi}{180}\cos x^{\circ}\).

How do you differentiate x sin x or e^x sin x?

Product rule \((uv)'=u'v+uv'\). So \(\dfrac{d}{dx}(x\sin x)=\sin x+x\cos x\) and \(\dfrac{d}{dx}(e^{x}\sin x)=e^{x}(\sin x+\cos x)\).

How do you find the gradient at an exact angle?

Evaluate the derivative there. For \(y=\sin x\), \(y'=\cos x\), so at \(x=\dfrac{\pi}{3}\) the gradient is \(\cos\dfrac{\pi}{3}=\dfrac12\).

How do you find stationary points of a trig function?

Set \(y'=0\) and solve on the domain. For \(y=\sin x+\cos x\), \(y'=\cos x-\sin x=0\Rightarrow\tan x=1\), so \(x=\dfrac{\pi}{4},\dfrac{5\pi}{4}\) on \([0,2\pi]\).

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The product rule

Differentiation of trigonometric functions

📖 Theory

How to differentiate a trigonometric function

Common pitfalls

Frequently asked questions

Theory