Derivative of xsinx
The derivative of xsinx is equal to xcosx + sinx. Differentiation is the process of determining the rate of change in a function with respect to the variable. We can evaluate the derivative of xsinx using the first principle of derivatives and the product rule of differentiation. To determine the differentiation of xsinx using the definition of the limits, we will use various trigonometric formulas and the specific formulas of the limits. We use the method of differentiation to find the maxima and minima of the function and slope of the curve at various points.
Further, in this article, we will compute the formula for the dertivative of xsinx using different methods of differentiation and trigonometric formulas. We will also solve various examples related to the differentiation of xsinx for a better understanding.
1.  What is Derivative of xsinx? 
2.  Derivative of xsinx Formula 
3.  Derivative of xsinx By First Principle of Derivatives 
4.  Derivative of xsinx Using the Product Rule 
5.  FAQs on Derivative of xsinx 
What is Derivative of xsinx?
The derivative of xsinx can be evaluated using the product rule of derivatives and the first principle of differentiation. The formula for the derivative of xsinx is given by, d(xsinx)/dx = xcosx + sinx. We use the derivative of sinx and x to arrive at the differentiation of xsinx. Also, the derivative of a function gives the rate of change of the function at a point. Differentiation of xsinx is nothing but the process of finding the derivative of xsinx. In the next section, let us go through the formula for the same.
Derivative of xsinx Formula
Now, the formula for the derivative of xsinx is given by d(xsinx)/dx = xcosx + sinx, where the differentiation is with respect to the variable x. In the next section, we will prove this formula of differentiation of xsinx using the first principle of derivatives, that is, the definition of limits. The image below shows the formula for the derivative of xsinx:
Derivative of xsinx By First Principle of Derivatives
Now that we know that the derivative of xsinx is equal to xcosx + sinx, we will now prove this formula using the first principle of derivatives, that is, the definition of limits. To find the derivative of xsinx, we take the limiting value as x approaches x + h. To simplify this, we set x = x + h, and we want to take the limiting value as h approaches 0. We will use the following formulas to prove the result:
 f'(x) = lim_{h→0}[f(x+h)  f(x)]/[(x+h)  x]
 sin A  sin B = 2 cos[(A+B)/2] sin[(AB)/2]
 lim_{x→0} (1/x) sin x = 1
d(xsinx)/dx = lim_{h→0}[(x+h)sin(x+h)  xsinx]/[(x+h)  x]
= lim_{h→0}[x sin(x+h) + h sin(x+h)  x sinx]/h
= lim_{h→0}[x (sin(x+h)  sinx) + h sin(x+h)]/h  [Rearranging the terms and taking x common and out.]
= lim_{h→0}[x (2 cos[(x+h+x)/2] sin[(x+hx)/2]) + h sin(x+h)]/h  [Using sin A  sin B formula]
= lim_{h→0}[x (2 cos[(2x+h)/2] sin[h/2]) + h sin(x+h)]/h
= lim_{h→0 }{ [x (2 cos[(2x+h)/2] sin[h/2])]/h + [h sin(x+h)]/h }
= lim_{h→0 }{ [x (cos[(2x+h)/2] sin[h/2])]/(h/2) + sin(x+h) }
= lim_{h→0 }{ [x (cos[(2x+h)/2] } × lim_{h→0 }[sin[h/2])]/(h/2)] + lim_{h→0 }sin(x+h)
= x cos (2x/2) × 1 + sin x  [Using the formula lim_{x→0} (1/x) sin x = 1]
= xcosx + sinx
Hence, we have proved that the derivative of xsinx is xcosx + sinx using the first principle of derivatives.
Derivative of xsinx Using the Product Rule
In this section, we will derive the derivative of xsinx using the product rule of derivatives. The product rule for the derivative of the function h(x) = uv is given as h'(x) = [uv]' = u'v + uv'. We will also use the formula for the derivative of sinx and the derivative of x to derive the derivative of xsinx. The formulas that we will use are:
 d(sinx)/dx = cosx
 dx/dx = 1
 [uv]' = u'v + uv'
Using the above formulas, we have
d(xsinx)/dx = (xsinx)'
= (x)' sinx + x (sinx)'
= 1 × sinx + x × cosx
= sinx + xcosx
Therefore, the derivative of xsinx is equal to xcosx + sinx using the product rule of differentiation.
Important Notes on Derivative of xsinx
 The derivative of xsinx is equal to xcosx + sinx.
 We can evaluate the differentiation of xsinx using the product rule and the first principle of derivatives.
 To find the second derivative of xsinx, we can differentiate its first derivative.
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Derivative of xsinx Examples

Example 1: Evaluate the derivative of e to the power xsinx.
Solution: To find the derivative of e to the power xsinx, we will use the chain rule of differentiation. We will use the formula for the derivative of the exponential function. We will also use the formula for the derivative of xsinx. Therefore, we have
d(e^{xsinx})/dx = e^{xsinx} × d(xsinx)/dx [Because d(e^{x})/dx = e^{x}]
= e^{xsinx} (xcosx + sinx)
Answer: The derivative of e to the power xsinx is equal to e^{xsinx} (xcosx + sinx).

Example 2: What is the second derivative of xsinx?
Solution: The second derivative of xsinx can be found by differentiating the first derivative of xsinx. Hence, we will differentiate the derivative of xsinx which is xcosx + sinx.
d^{2}(xsinx)/dx^{2} = d(xcosx + sinx)/dx
= d(xcosx)/dx + d(sinx)/dx
= (dx/dx) cosx + x d(cosx)/dx + cosx
= cosx  xsinx + cosx
= 2 cosx  xsinx
Answer: The second derivative of xsinx is equal to 2 cosx  xsinx.
FAQs on Derivative of xsinx
What is the Derivative of xsinx in Math?
The derivative of xsinx is given by d(xsinx)/dx = xcosx + sinx, where the differentiation is with respect to the variable x. We can find this derivative using different methods of differentiation.
What is the Formula for the Derivative of xsinx?
The formula for the derivative of xsinx is given by d(xsinx)/dx = xcosx + sinx. We can also write this formula as (xsinx)' = xcosx + sinx where the prime symbol indicates the derivative of the function.
How to Find the Derivative of xsinx?
We can find the derivative of xsinx using various methods of differentiation such as the product rule of differentiation and the first principle of derivatives. For using the product rule, we can consider x as the first function and sinx as the second function.
What is the Second Derivative of xsinx?
The second derivative of xsinx is 2cosx  xsinx which can be determined by differentiating the first derivative of xsinx. The first derivative of xsinx is given by d(xsinx)/dx = xcosx + sinx.
What is the Derivative of xsinx + cosx?
The derivative of xsinx + cosx is given by, (xsinx + cosx)' = (xsinx)' + (cosx)' = xcosx + sinx  sinx = xcosx. Hence the derivative of xsinx + cosx is equal to xcosx.
What is the 35^{th} Derivative of xsinx?
We know that the first derivative of xsinx is xcosx + sinx. Differentiating this further for y = xsinx, we have:
 y'' = 2cosx  xsinx
 y''' = 3sinx  xcosx
 y^{(4)} = cosx + xsinx
Continuing like this, we have the 35^{th} derivative of xsinx to be equal to 35sinx  xcosx.
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