How do you differentiate #p(y) = y^2sin(y)# using the product rule?

Answer 1

Scarlett Allison

#p'(y)=2ysin(y)+y^2cos(y)#

According to the product rule:

#p'(y)=sin(y)d/dy[y^2]+y^2d/dy[sin(y)]#

Find each derivative separately.

#d/dy[y^2]=2y#

#d/dy[sin(y)]=cos(y)#

Plug them back in:

#p'(y)=2ysin(y)+y^2cos(y)#

If you want to factor:

#p'(y)=y(2sin(y)+ycos(y))#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Samantha Adkison

To differentiate ( p(y) = y^2\sin(y) ) using the product rule:

Let ( f(y) = y^2 ) and ( g(y) = \sin(y) ).

The product rule states:

[ (f \cdot g)' = f' \cdot g + f \cdot g' ]

Where ( f' ) and ( g' ) denote the derivatives of ( f ) and ( g ) respectively.

Now, find the derivatives:

[ f'(y) = 2y ]

[ g'(y) = \cos(y) ]

Apply the product rule:

[ p'(y) = (2y) \cdot \sin(y) + y^2 \cdot \cos(y) ]

[ \boxed{p'(y) = 2y\sin(y) + y^2\cos(y)} ]

This is the derivative of ( p(y) = y^2\sin(y) ) using the product rule.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.