If y = f(x)g(x), then dy/dx = f‘(x)g‘(x). If it is true, explain your answer. If false, provide a counterexample. True or False?

Answer 1

False

Take #f(x)=x^2+1,g(x)=x# then
#f(x)g(x)=x^3+x# then #f'(x)*g'(x)=2x# But
#(f(x)g(x))'=3x^2+1# It must be #(f(x)g(x))'=f'(x)*g(x)+f(x)*g'(x)#
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Answer 2

The statement is false .

The product rule provides the correct formulation:

#y = f(x)g(x) => y = f(x)g'(x) + f'(x)g(x) #

We can readily disprove the given statement:

Consider:

#f(x)=x# and #g(x)=x#
Then differentiating wrt #x# we have:
#f'(x)=1# and #g'(x)=1 => f'(x)g'(x)=1#
And #y=f(x)g(x) =x^2 => dy/dx = 2x != f'(x)g'(x)#

And so By counterexample, the statement is false .

In fact the product rule provides the correct formulation:

#y = f(x)g(x) => y = f(x)g'(x) + f'(x)g(x) #
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Answer 3

False. The correct formula for the product rule is ( \frac{{dy}}{{dx}} = f'(x)g(x) + f(x)g'(x) ). This rule states that when differentiating the product of two functions, ( f(x) ) and ( g(x) ), with respect to ( x ), the result is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

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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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