The derivative of sinx/x is equal to (xcosx -sinx)/x^{2}. Here we will learn how to differentiate sinx/x with respect to x.

The derivative formula of sinx/x is given by

$\dfrac{d}{dx}(\dfrac{\sin x}{x})$ = $\dfrac{x\cos x -\sin x}{x^2}$.

## Find the Derivative of sinx/x

As sinx/x is a quotient function, to find the derivative of sinx/x we will use the quotient rule of derivatives. By the rule, the derivative of $\dfrac{f}{g}$ is given by

$\big(\dfrac{f}{g}\big)’$ = $\dfrac{gf’-fg’}{g^2}$

where $’$ denotes the first order derivative with respect to x.

Put f=sinx and g=x, the derivative of sinx/x is equal to

$\big(\dfrac{\sin x}{x}\big)’$ = $\dfrac{x \cdot (\sin x)’-\sin x \cdot (x)’}{x^2}$

= $\dfrac{x \cdot \cos x-\sin x \cdot 1}{x^2}$ as the derivative of sinx is cosx and the derivative of x is 1.

= $\dfrac{x \cdot \cos x-\sin x \cdot 1}{x^2}$

= $\dfrac{x\cos x -\sin x}{x^2}$.

So the derivative of sinx/x is (xcosx -sinx)/x^{2} and it is obtained by the quotient rule of derivatives.

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## Question-Answer on Derivative of sinx/x

**Question:** What is the derivative of sinx/x at x=1?

*Answer:*

From above, the derivative of sinx/x at x=1 is equal to

[(sinx/x)$’$] = [(xcosx -sinx)/x^{2}]_{x=1}

= (1 ⋅ cos1 -sin1)/1^{2}

= cos1 -sin1.

So the derivative of sinx/x at the point x=1 is equal to cos1 -sin1.

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## FAQs

**Q1: What is the derivative of sinx/x?**

Answer: The derivative of sinx/x, denoted by d/dx(sinx/x), is equal to (xcosx -sinx)/x^{2}.