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Question: How do you differentiate \(({x^2})(\sin x)?\)...

How do you differentiate (x2)(sin⁡x)?({x^2})(\sin x)?

Explanation

Solution

In these types of questions, we need to use the formula of product rule which is given below as: f′(x)=g′(x)h(x)+h′(x)g(x)f'(x) = g'(x)h(x) + h'(x)g(x) . Find each and every term of the formula and put the values in it and get the solution as we require.

Formula used: 1) Product rule: f′(x)=g′(x)h(x)+h′(x)g(x)f'(x) = g'(x)h(x) + h'(x)g(x)
2) d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}
3) d(sin⁡x)dx=cos⁡x\dfrac{{d(\sin x)}}{{dx}} = \cos x

Complete step-by-step solution:
We have given an expression as below:
We can consider the given function to be f(x)f(x) .
Let f(x)=(x2)(sin⁡x)f(x) = ({x^2})(\sin x)
Now, we need to define given function in the form of two new functions as g(x)g\left( x \right) and h(x)h\left( x \right)
So, we get the following expression:
Then, f(x)=g(x)×h(x)f(x) = g(x) \times h(x)
The formula for the derivative of this function is product rule and it is mentioned below:
f′(x)=(g′(x)×h(x))+(h′(x)×g(x))......(A)f'(x) = (g'(x) \times h(x)) + (h'(x) \times g(x))......(A)
The derivative of g(x)g(x) or x2{x^2} is equal to g′(x)=2×x2−1=2x.........(1)g'(x) = 2 \times {x^{2 - 1}} = 2x.........(1) Using the formula (2)(2)
The derivative of h(x)h(x) or sin⁡x\sin x
h′(x)=cos⁡x..........(2)h'(x) = \cos x..........(2) …. Using the formula (3)(3)
Applying the product rule:
f′(x)=(g′(x)×h(x))+(h′(x)×g(x))f'(x) = (g'(x) \times h(x)) + (h'(x) \times g(x))
Putting the values we are getting in the equations (1)(1) and (2)(2) in the equation (A)(A) we can have,
f′(x)=(2x(sin⁡x))+(x2(cos⁡x))f'(x) = (2x(\sin x)) + ({x^2}(\cos x))
Solve the brackets by multiplying the terms we get,
f′(x)=2xsin⁡x+x2cos⁡xf'(x) = 2x\sin x + {x^2}\cos x
Hence, the derivative of y=(x2)(sin⁡x)y = ({x^2})(\sin x)
dydx=\dfrac{{dy}}{{dx}} = y′=2xsin⁡x+x2cos⁡xy' = 2x\sin x + {x^2}\cos x
Therefore, we differentiated the given function which we required.

Note: When finding the derivative of any function, in most cases, we have to use the product rule of differentiation. We have given the formula as below:
f(′x)=g′(x)h(x)+h′(x)g(x)f('x) = g'(x)h(x) + h'(x)g(x)
This can be seen in the problem above where we had to use it.
-Differentiating is the method of finding the derivative of a function and finding the rate of change of function with respect to one.
-To differentiate something means to take the derivative of that value. Taking the derivative of a function is the same as finding the slope at any point, so differentiating is just similar to finding the slope.