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Trigonometric derivatives

Derivative of cos(x): The Negative Sine Sign Explained

Cosine begins at a maximum and initially decreases, so its derivative must be negative near zero. The difference quotient makes that sign precise.

Direct answer

ddx[cos⁡x]=−sin⁡x\frac{\mathrm{d}}{\mathrm{d}x}\left[\cos x\right] = -\sin x

The standard formula assumes radians; cos(g(x)) also requires g′(x).

When to use derivative of cos(x)

Choose the method from the top-level mathematical structure before manipulating symbols. Conditions are part of the task, not optional notes.

Ask first 1

Which structure controls the method?

What to do

Use the cosine rule for cos(g(x)), then inspect whether g(x) is composite.

Ask first 2

Which condition can change the answer?

What to do

The standard formula assumes radians; cos(g(x)) also requires g′(x).

Ask first 3

How should the result be checked?

What to do

Check the negative sign near zero and confirm the original angle remains inside sine.

Track the sign in the difference quotient

Expand cos(x+h)=cos(x)cos(h)−sin(x)sin(h).

Subtract cos(x), divide by h, and separate the two standard limits.

The cosine term tends to zero and the sine ratio tends to one, leaving −sin(x).

For cos(g(x)), multiply −sin(g(x)) by g′(x).

1
cos⁡(x+h)=cos⁡x cos⁡h−sin⁡x sin⁡h\cos\left(x + h\right) = \cos x\,\cos h - \sin x\,\sin h
2
lim sin(h)/h=1
3
[cos(g(x))]'=−sin(g(x))g'(x)

Worked examples

ddx[cos⁡x]\frac{\mathrm{d}}{\mathrm{d}x}\left[\cos x\right]
−sin⁡x-\sin x

The negative sign is part of the rule.

ddx[cos⁡(5 x)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\cos\left(5\,x\right)\right]
−5 sin⁡(5 x)-5\,\sin\left(5\,x\right)

The inner derivative is 5.

ddx[cos⁡(x3+1)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\cos\left(x^{3} + 1\right)\right]
−3 x2 sin⁡(x3+1)-3\,x^{2}\,\sin\left(x^{3} + 1\right)

Keep x³+1 inside sine.

Common mistakes

  • • Dropping the negative sign.
  • • Replacing the original angle.
  • • Forgetting the inner derivative.
  • • Using the sine rule for cosine.

Error clinic: locate the first invalid step

Incorrect attempt

[cos(x)]'=sin(x)

Why it fails

The direction of change is reversed.

Correction

[cos(x)]'=−sin(x)

Incorrect attempt

[cos(5x)]'=−sin(5x)

Why it fails

The inner derivative 5 is missing.

Correction

[cos(5x)]'=−5sin(5x)

Practice the same idea with a changed structure

Name the rule and its conditions before revealing the reference answer. Explain every sign and factor.

ddx[cos⁡(5 x)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\cos\left(5\,x\right)\right]

Check before solving

The inner derivative is 5.

Reveal the reference answer

Reference answer

−5 sin⁡(5 x)-5\,\sin\left(5\,x\right)
ddx[cos⁡(x3+1)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\cos\left(x^{3} + 1\right)\right]

Check before solving

Keep x³+1 inside sine.

Reveal the reference answer

Reference answer

−3 x2 sin⁡(x3+1)-3\,x^{2}\,\sin\left(x^{3} + 1\right)

Try the worked examples in the calculator

Each link opens the matching calculator with the expression, variable and required conditions already filled in.

Derivadas:cos⁡x\cos xDerivadas:cos⁡(x3+1)\cos\left(x^{3} + 1\right)
Open the calculator without a preset: Derivadas

Frequently asked questions

Why is the derivative negative?+

Cosine decreases immediately to the right of zero, and the quotient preserves that direction.

What is the second derivative?+

Differentiate −sin(x) to obtain −cos(x).

What changes for cos(−x)?+

The chain rule agrees with the identity cos(−x)=cos(x), so the derivative remains −sin(x).

Conclusion

Cosine differentiates to negative sine. Preserve the sign and multiply by any inner derivative.

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