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Question

Question: How do you find the derivative of \[\dfrac{{\sin x}}{x}\]?...

How do you find the derivative of sinxx\dfrac{{\sin x}}{x}?

Explanation

Solution

In the given question, we have been given a trigonometric expression. It is a combination of two trigonometric functions being divided. We have to find the derivative of this trigonometric expression. To do that, we apply the quotient rule of differentiation on the trigonometric functions.

Formula Used:
We are going to use the formula of quotient rule of differentiation, which is:
(uv)=uvuvv2{\left( {\dfrac{u}{v}} \right)^\prime } = \dfrac{{u'v - uv'}}{{{v^2}}}

Complete step by step solution:
The given trigonometric expression is sinxx\dfrac{{\sin x}}{x}.
We are going to use the formula of quotient rule of differentiation, which is:
(uv)=uvuvv2{\left( {\dfrac{u}{v}} \right)^\prime } = \dfrac{{u'v - uv'}}{{{v^2}}}
Here, u=sinxu = \sin x and v=xv = x
Substituting the values into the formula, we get,
(sinxx)=(sinx)xsinx(x)x2{\left( {\dfrac{{\sin x}}{x}} \right)^\prime } = \dfrac{{{{\left( {\sin x} \right)}^\prime }x - \sin x{{\left( x \right)}^\prime }}}{{{x^2}}}
Now, (sinx)=cosx{\left( {\sin x} \right)^\prime } = \cos x and (x)=1{\left( x \right)^\prime } = 1
Hence,

=xcosxsinxx = \dfrac{{x\cos x - \sin x}}{x}

Note:
In the given question, we had been given a trigonometric expression. It was a division of two trigonometric functions. We had to find the derivative of this trigonometric expression. We solved this question by applying the quotient rule of differentiation on the trigonometric functions and by using the standard relation values. Some students make mistakes when they do not remember the standard result values and get confused with it. So, it is important to know these values and how to use these values for the given question.