Derivative of sin x: Proof, Examples and Trig Derivatives

The derivative of ( sin x ) is ( cos x ) (with x in radians). By the chain rule, ( frac{d}{dx}sin u = cos ucdot u’ ), so ( frac{d}{dx}sin(3x) = 3cos(3x) ) and ( frac{d}{dx}sin^2 x = 2sin xcos x = sin 2x ). The derivative of ( cos x ) is ( -sin x ).

The derivative formula for the sine function is:

$$frac{d}{dx}sin x = cos x$$

This holds when you measure x in radians. Geometrically, the height of the cosine curve gives the exact slope of the sine curve at any given point:

• At ( x = 0 ), ( sin(0) = 0 ) and ( cos(0) = 1 ). The sine curve crosses the origin with a slope of 1.
• At ( x = frac{pi}{2} ), ( sin(frac{pi}{2}) = 1 ) and ( cos(frac{pi}{2}) = 0 ). The sine curve reaches its peak, where the tangent line is horizontal with a slope of 0.
• At ( x = pi ), ( sin(pi) = 0 ) and ( cos(pi) = -1 ). The sine curve crosses the axis heading downward with a slope of -1.
• At ( x = frac{3pi}{2} ), ( sin(frac{3pi}{2}) = -1 ) and ( cos(frac{3pi}{2}) = 0 ). The sine curve reaches its trough with a slope of 0.

All six trig derivatives

Function	Derivative
( sin x )	( cos x )
( cos x )	( -sin x )
( tan x )	( sec^2 x )
( cot x )	( -csc^2 x )
( sec x )	( sec xtan x )
( csc x )	( -csc xcot x )

A simple memory aid keeps the signs straight: the derivatives of all three “co-” functions (cosine, cotangent, and cosecant) begin with a minus sign. Also, you can derive all secondary trigonometric derivatives directly from sine and cosine using quotient and reciprocal rules. Read detailed guides for specific functions here: derivative of tan x, derivative of sec x, and derivative of cos²x.

Proof from the limit definition

To prove why ( frac{d}{dx}sin x = cos x ), you start from the limit definition of the derivative:

$$frac{d}{dx}sin x = lim_{hto0}frac{sin(x + h) - sin x}{h}$$

Step 1: expand using angle addition. Apply the angle-addition identity (see trig identities), ( sin(x + h) = sin xcos h + cos xsin h ). Next, substitute and group terms by ( sin x ) and ( cos x ):

$$lim_{hto0}left[ sin xcdotfrac{cos h - 1}{h} + cos xcdotfrac{sin h}{h} right]$$

Step 2: apply two standard trigonometric limits.

$$lim_{hto0}frac{sin h}{h} = 1, qquad lim_{hto0}frac{cos h - 1}{h} = 0$$

You prove the first limit using the squeeze theorem in the limit of sin x / x guide. To prove the second limit algebraically, multiply the numerator and denominator by the conjugate ( (cos h + 1) ):

$$lim_{hto0}frac{cos h - 1}{h}cdotfrac{cos h + 1}{cos h + 1} = lim_{hto0}frac{cos^2 h - 1}{h(cos h + 1)} = lim_{hto0}frac{-sin^2 h}{h(cos h + 1)}$$

Splitting the terms gives:

$$lim_{hto0}left( -frac{sin h}{h}cdotfrac{sin h}{cos h + 1} right) = -1cdotfrac{0}{1 + 1} = -1cdot 0 = 0$$

Step 3: calculate the final result. Next, substitute these two limits back into the expression:

$$sin xcdot 0 + cos xcdot 1 = cos x$$

Why radians matter

The limit ( lim_{hto0}frac{sin h}{h} = 1 ) relies directly on measuring angle h in radians, where arc length on a unit circle matches the central angle. However, if you work in degrees, ( lim_{hto0}frac{sin h^circ}{h} = frac{pi}{180} ). Differentiating ( sin(x^circ) ) in degrees yields an extra constant factor:

$$frac{d}{dx}sin(x^circ) = frac{pi}{180}cos(x^circ)$$

Calculus uses radians exclusively so derivative formulas stay clean and free of unnecessary scaling conversion factors.

Derivative of cos x and higher-order derivatives

Applying the limit definition with ( cos(x + h) = cos xcos h – sin xsin h ) yields the derivative of cosine:

$$frac{d}{dx}cos x = -sin x$$

Repeated differentiation follows a repeating four-step cycle:

1. First derivative: ( frac{d}{dx}sin x = cos x )
2. Second derivative: ( frac{d^2}{dx^2}sin x = -sin x )
3. Third derivative: ( frac{d^3}{dx^3}sin x = -cos x )
4. Fourth derivative: ( frac{d^4}{dx^4}sin x = sin x )

Because the cycle repeats every four derivatives, you find any nth derivative by dividing n by 4 and taking the remainder. For example, to find the 100th derivative of ( sin x ), divide 100 by 4. The remainder is 0, so the 100th derivative equals ( sin x ). For the 47th derivative, dividing 47 by 4 leaves a remainder of 3, corresponding to the third step: ( -cos x ).

Chain rule examples

When the input is a function ( u(x) ) rather than plain x, apply the chain rule: ( frac{d}{dx}sin(u) = cos(u)cdot u’ ).

Example 1: sin(3x). The inner function is ( u = 3x ) with derivative ( u’ = 3 ).

$$frac{d}{dx}sin(3x) = cos(3x)cdot 3 = 3cos(3x)$$

Example 2: sin(x²). The inner function is ( u = x^2 ) with derivative ( u’ = 2x ).

$$frac{d}{dx}sin(x^2) = cos(x^2)cdot 2x = 2xcos(x^2)$$

Example 3: sin²x. Write this as ( (sin x)^2 ). The outer function is the square power, and the inner function is ( sin x ):

$$frac{d}{dx}sin^2 x = 2(sin x)^1cdotcos x = 2sin xcos x = sin 2x$$

Example 4: sin⁵(2x). Write this as ( (sin(2x))^5 ). You apply two layers of the chain rule:

$$frac{d}{dx}sin^5(2x) = 5(sin(2x))^4cdotcos(2x)cdot 2 = 10sin^4(2x)cos(2x)$$

Example 5: √(sin x). Express as ( (sin x)^{1/2} ):

$$frac{d}{dx}(sin x)^{1/2} = frac12(sin x)^{-1/2}cos x = frac{cos x}{2sqrt{sin x}}$$

Example 6: sin(cos x). The outer function is sine, and the inner function is cosine:

$$frac{d}{dx}sin(cos x) = cos(cos x)cdot(-sin x) = -sin xcos(cos x)$$

Product and quotient examples

Combine basic derivative rules with trigonometric differentiation to solve products and quotients.

Example 7: x² sin x. Apply the product rule ( (uv)’ = u’v + uv’ ):

$$frac{d}{dx}left(x^2sin xright) = (2x)(sin x) + (x^2)(cos x) = 2xsin x + x^2cos x$$

Example 8: sin x / x. Apply the quotient rule ( left(frac{u}{v}right)’ = frac{u’v – uv’}{v^2} ):

$$frac{d}{dx}frac{sin x}{x} = frac{(cos x)(x) - (sin x)(1)}{x^2} = frac{xcos x - sin x}{x^2}$$

Example 9: tan x from sine and cosine. Express tangent as ( tan x = frac{sin x}{cos x} ) and differentiate using the quotient rule:

$$frac{d}{dx}tan x = frac{(cos x)(cos x) - (sin x)(-sin x)}{cos^2 x} = frac{cos^2 x + sin^2 x}{cos^2 x}$$

Apply the Pythagorean identity ( cos^2 x + sin^2 x = 1 ):

$$frac{d}{dx}tan x = frac{1}{cos^2 x} = sec^2 x$$

Try it: derivative calculator

Enter any trig expression to see the derivative with every step. The example below processes ( sin(3x^2) ), yielding ( 6xcos(3x^2) ).

The full derivative calculator handles higher-order derivatives; see the second derivative calculator to track how sine and cosine alternate under repeated steps.

Applications

• Tangent lines and linear approximations. At ( x = 0 ), ( sin(0) = 0 ) and the derivative slope is ( cos(0) = 1 ). The tangent line equation is ( y = x ). This forms the foundation of linear approximation, where ( sin x approx x ) for small x values.
• Simple harmonic motion. An oscillating spring’s position ( x(t) = Asin(omega t) ) yields velocity ( v(t) = Aomegacos(omega t) ) and acceleration ( a(t) = -Aomega^2sin(omega t) = -omega^2 x(t) ).
• Optimization. To find extreme values of ( f(x) = sin x + cos x ), set the derivative to zero: ( cos x – sin x = 0 ), meaning ( tan x = 1 ). On ( [0, 2pi] ), this gives critical points at ( x = frac{pi}{4} ) and ( x = frac{5pi}{4} ), with a maximum value of ( sqrt{2} ).

Common mistakes

• Sign errors on cosine. Remember that ( frac{d}{dx}sin x = cos x ) remains positive, while ( frac{d}{dx}cos x = -sin x ) takes a minus sign.
• Forgetting the inner derivative. Writing ( frac{d}{dx}sin(5x) = cos(5x) ) leaves out the inner rate of change. You must multiply by 5 to get ( 5cos(5x) ).
• Confusing sin²x with sin(x²). The expression ( sin^2 x ) means ( (sin x)^2 ), yielding ( 2sin xcos x ). The expression ( sin(x^2) ) has a squared argument, yielding ( 2xcos(x^2) ).
• Using degree inputs directly. Standard derivative formulas require radians. Convert degree values by multiplying by ( frac{pi}{180} ) before differentiating.

When standard trig differentiation needs modification

The standard derivative formula ( frac{d}{dx}sin x = cos x ) applies to real-valued functions defined on continuous domains in radians. Modify your approach in these scenarios:

• Degree inputs. When working in degrees, multiply your derivative result by ( frac{pi}{180} ).
• Non-differentiable absolute value corners. Functions like ( y = |sin x| ) feature sharp points at integer multiples of ( pi ) where two-sided derivatives do not exist.
• Inverse trigonometric functions. Differentiating ( arcsin x ) yields an algebraic expression rather than a trigonometric function: ( frac{d}{dx}arcsin x = frac{1}{sqrt{1 – x^2}} ).

Practice problems

Find the derivative of each function.

1. ( y = 4sin x – 3cos x )
2. ( y = sin(7x) )
3. ( y = cos(x^3) )
4. ( y = xsin x )
5. ( y = sin^3 x )
6. ( y = sin xcos x )

Detailed solutions

1. Problem 1: Apply basic derivative rules directly: ( y’ = 4(cos x) – 3(-sin x) = 4cos x + 3sin x ).
2. Problem 2: Apply the chain rule with inner derivative 7: ( y’ = cos(7x)cdot 7 = 7cos(7x) ).
3. Problem 3: Apply the chain rule on cosine with inner derivative ( 3x^2 ): ( y’ = -sin(x^3)cdot 3x^2 = -3x^2sin(x^3) ).
4. Problem 4: Apply the product rule with ( u = x ) and ( v = sin x ): ( y’ = (1)(sin x) + (x)(cos x) = sin x + xcos x ).
5. Problem 5: Apply the power chain rule on ( (sin x)^3 ): ( y’ = 3(sin x)^2cdot cos x = 3sin^2 xcos x ).
6. Problem 6: Apply the product rule: ( y’ = (cos x)(cos x) + (sin x)(-sin x) = cos^2 x – sin^2 x = cos(2x) ). Alternatively, rewrite ( sin xcos x = frac12sin(2x) ), whose derivative is ( frac12(2cos(2x)) = cos(2x) ).

Frequently asked questions

What is the derivative of sin x?

The derivative of ( sin x ) is ( cos x ), provided x is measured in radians.

What is the derivative of sin(2x)?

The derivative of ( sin(2x) ) is ( 2cos(2x) ). You get this by applying the chain rule, multiplying ( cos(2x) ) by the inner derivative of 2x.

What is the derivative of sin²x?

The derivative of ( sin^2 x ) is ( 2sin xcos x ), which equals ( sin(2x) ) by the double-angle identity.

What is the integral of sin x?

The integral is ( int sin x,dx = -cos x + C ), because differentiating ( -cos x ) returns positive ( sin x ).

Why is the derivative of sin x equal to cos x?

It equals ( cos x ) because applying the angle-addition formula to the limit definition yields ( sin xcdot 0 + cos xcdot 1 ), driven by the core limits ( lim_{hto0}frac{sin h}{h} = 1 ) and ( lim_{hto0}frac{cos h – 1}{h} = 0 ).

Further reading

• Derivatives of Trig Functions (Paul’s Online Math Notes) — proofs of the sine and cosine derivatives and the rest of the table.
• 3.5 Derivatives of Trigonometric Functions (OpenStax Calculus Volume 1) — derivations, higher-order derivatives and applications.