How do you differentiate #f(x)= 7xsinx# using the product rule?

Answer 1

#f'(x) = 7sinx + 7xcosx#

According to the product rule:

#f(x) = g(x)h(x) -> f'(x) = g'(x)h(x)+g(x)h'(x)#
So for #f(x) = 7xsinx# we can take:
#g(x) = 7x# and #h(x) = sinx#
Thus #g'(x) = 7# and #h'(x) = cosx#

Adding this in place of the product rule:

#f'(x) = 7sinx + 7xcosx#
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Answer 2

To differentiate the function ( f(x) = 7x \sin(x) ) using the product rule:

Let ( u(x) = 7x ) and ( v(x) = \sin(x) ).

Using the product rule, the derivative of ( f(x) ) with respect to ( x ) is:

[ f'(x) = u(x)v'(x) + v(x)u'(x) ]

where ( u'(x) ) and ( v'(x) ) are the derivatives of ( u(x) ) and ( v(x) ) respectively.

First, find ( u'(x) ) and ( v'(x) ):

( u'(x) = 7 ) (derivative of ( 7x ) with respect to ( x ))

( v'(x) = \cos(x) ) (derivative of ( \sin(x) ) with respect to ( x ))

Now, plug these into the product rule formula:

[ f'(x) = (7x)(\cos(x)) + (\sin(x))(7) ]

So, the derivative of ( f(x) ) with respect to ( x ) is:

[ f'(x) = 7x\cos(x) + 7\sin(x) ]

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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.

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.

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.

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.

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