How do find the derivative of #y = x^2 sinx#?

Question

Cameron Alexander · Accepted Answer

#dy/dx = 2xsin(x) + x^2cos(x)#

The product rule states the following and can be used to determine the derivative:

If #h(x) = f(x)g(x)#, then #h'(x) = f'(x)g(x) + f(x)g'(x)#

In this instance,

#h(x) = x^2sinx#

#f(x) = x^2#

#g(x) = sinx#

#f'(x) = 2x#

#g'(x) = cosx#

Enter those values into our product rule definition to obtain

#dy/dx = h'(x) = 2xsin(x) + x^2cos(x)#

Elijah Abram · Answer

undefined To find the derivative of $ y = x^2 \sin(x) $, you can use the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Applying the product rule to this function, the derivative is $ y' = 2x \sin(x) + x^2 \cos(x) $.

Emilia Aguilar · Answer

undefined To find the derivative of y = x^2 sin(x), you can use the product rule of differentiation. The product rule states that if you have two functions u(x) and v(x), then the derivative of their product u(x) * v(x) is given by:

(uv)' = u'v + uv'

Applying this rule to y = x^2 sin(x), where u(x) = x^2 and v(x) = sin(x), we get:

y' = (x^2)'sin(x) + x^2(sin(x))'

Now, find the derivatives of each term:

(u)' = 2x
(v)' = cos(x)

Substituting these derivatives into the equation, we get:

y' = (2x)sin(x) + x^2(cos(x))

So, the derivative of y = x^2 sin(x) is y' = 2x sin(x) + x^2 cos(x).