**Product Rule** in differentiation applies to functions that are a multiple of each other. For example, assume we have two functions h and g and suppose further that f=h\cdot g. Then, If we are asked to find the derivative f, then we have to differentiate the product of the functions h and g of course. The question is, how do we do this? As for now, you maybe thinking and guessing for the answer. Some of you might even think that the derivative of h\cdot g is h'\cdot g'. Well, you have a topnotch guess. Do you know why? There is a story that even Leibniz himself, one of the Calculus' discoverer, got that same guess initially. But I have to tell you that your guess is wrong. Such error is common when it comes to this topic in derivative. Even the great Leibniz!
So much for history. To make things easier, let me now show you how to find the derivative of the f.

If h and g are differentiable at x, and f(x)=h(x)\cdot g(x), |

Find the derivative of f when f(x)=(2x)\cdot x^{2}.

We know both the derivative of 2x and 2x^{2}, so let's differentiate directly using the theorem above.

So,

f'(x)=2x\cdot (2x) +x^{2}\cdot 2=4x^{2}+2x^{2}=6x^{2}

Thus,

f'(x)=6x^{2}

Find the derivative of the function f(x)=x\sin{x}.

We know both the derivative of x and \sin{x}, so let's differentiate directly using the theorem above.

So,

f'(x)=x\cdot \cos{x} +1\cdot \sin{x}=x\cos{x} + \sin{x}

Thus,

f'(x)=x\cos{x} + \sin{x}

Using Product Rule find the derivative of f when f(x)=x^2\cdot \sin{x}\cos{x}.

We have a function which consists of three products, thus we may apply product rule twice.

f'(x)=x^2\cdot D_x[\sin{x}\cos{x}] + 2x \cdot [\sin{x}\cos{x}]

Applying again the product rule on the D_x[\sin{x}\cos{x}] term, we get

f'(x)= x^2\cdot [-\sin^2{x}+\cos^2{x}] + 2x \cdot [\sin{x}\cos{x}]

Thus,

f'(x)=2x \cdot [\sin{x}\cos{x}]+x^2\cos^2{x} - x^2\sin^2{x}

Find the derivative of the function f when f(x)=\frac{1}{2}xe^{2x}.

By virtue of the product rule, we proceed with direct differentiation.

f'(x)=\frac{1}{2}x\cdot 2e^{2x}+\frac{1}{2}e^{2x}\cdot (1)

Thus,

f'(x)=e^{2x}[ x+\frac{1}{2}

Solve for the derivative of f(x)=3y^2x^3.

This is just direct differentiation, no more product rule involved.

f'(x)=3y^2\cdot D_x x^3

=3y^2 \cdot \frac{1}{3}x^2

Thus,

f'(x)=y^2x^{2}

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Page last modified on June 30, 2011