The derivative of the mod cosx is equal to (- sinx cosx)/|cosx|. Here let us learn how to differentiate mod cosx, that is, how to find d/dx(|cosx|).

The derivative formula of mod cosx is given as follows.

$\boxed{\dfrac{d}{dx}(|\cos x|)=\dfrac{-\sin x \cos x}{|\cos x|}}$,

provided that cosx is non-zero.

The following formula is very useful to find the derivative of mod cosx.

$\boxed{\dfrac{d}{dx}(|x|)=\dfrac{x}{|x|}}$ for $x \neq 0$ **…( ∗)**

For details, visit derivative of mod x.

## Derivative of modulus of cosx

**Question:** Find the derivative of |cosx|.

**Answer:**

Put z = cosx.

Differentiating, dz/dx = -sinx **…(∗ ∗)**

Now, by the chain rule of derivatives, we have that

$\dfrac{d}{dx}(|\cos x|)$ = $\dfrac{d}{dz}(|z|) \times \dfrac{dz}{dx}$

= $\dfrac{z}{|z|} \times (-\sin x)$ by **( ∗)** and

**(**

**∗**∗)= $\dfrac{-\sin x \cos x}{|\cos x|}$ as z=cosx.

So the derivative of mod cosx by the chain rule is equal to (-sinx cosx)/|cos x|.

## Question-Answer

**Question 1:** Find the derivative of mod cos2x.

**Answer:**

Let t=2x.

⇒ dt/dx = 2

By the chain rule of derivatives,

$\dfrac{d}{dx}(|\cos 2x|)$ = $\dfrac{d}{dt}(|\cos t|) \times \dfrac{dt}{dx}$

= $\dfrac{-\sin t \cos t}{|\cos t|} \times 2$ by the above derivative formula of |cosx|.

= $\dfrac{-2\sin t \cos t}{|\cos t|}$

= $\dfrac{-\sin 2t}{|\cos t|}$ by the identity sin2θ=2sinθ cosθ.

= $\dfrac{-\sin 4x}{|\cos 2x|}$ as t=2x.

So the derivative of mod cos2x is equal to (-sin4x) / |cos 2x|.

**ALSO READ:**

Derivative of mod sinx | Derivative of root sinx

Derivative of root x | Derivative of root e^{x}

## FAQs

**Q1: **If y=|cosx|, then find dy/dx?

**Answer:** If y=|cosx|, then dy/dx = (-sinx cosx) / |cos x|.