# The Product Rule - Problem 2

By the product rule, the derivative of a product of functions f(x)g(x) is f(x)g'(x) + f'(x)g(x). So, in order to calculate the derivative of a product of functions, calculate the derivatives of the individual functions. Then, simply multiply the derivative of the second function to the first function, and add the derivative of the first function multiplied by the second function.

For example, if your function h(x)=x^{2}ln(x), let f(x)=x^{2} and g(x)=ln(x). We know by the rules for deriving power functions and logarithmic functions that f'(x)=2x and g'(x)=1/x. h'(x)=f(x)g'(x) + f'(x)g(x). So, h'(x)=x^{2}(1/x) + 2xln(x) = x + 2xln(x).

Sometimes, when given the multiplication of two long polynomial functions, it may be easier to use the product rule, instead of trying to expand the polynomial.

So we're talking about the product rule. Remember that says the derivative of the product of two functions is the first times the derivative of the second, plus the second times the derivative of the first.

So let's apply that to these two problems. Here I'm asked to differentiate x² times 2 to the x. That's going to give me dy/dx equals the first x² times the derivative of the second, 2 to the x, plus the second 2to the x times the derivative of the first, x².

So here I have to remember what the derivative of 2 to the x is. Remember I did this deliberately. X² and 2 to the x look very similar. This is a power function. A power function is a function where x is in the base. An exponential function is a function where the x is in the exponent. You differentiate these differently. The derivative of 2 to the x is ln2 times 2 to the x, so times x².

Then over here, the derivative of x² is 2x. So this is 2 to the x times 2x. Then also it's easy to get into trouble, because if your x's start to float up, one thing can become another. This is pretty good. What I can do, is I can factor out anything that's common. I see the 2 to the x's are common here. There's also a common factor of x, but I'll just factor out the 2 to the x for now. We have ln2 times x² plus 2x. That's our derivative.

Let's take a look at this guy. Now this function is a product of two polynomials. If you multiplied it through, you'd probably just get one big polynomial. You could differentiate it that way. You can also bypass that step, because it would be a lot of Algebra. You can bypass that and differentiate it using the product rule.

So dy/dx equals the first times the derivative of the second. Instead of writing it I'm just going to differentiate it. To to be 2x minus 3. Plus the second, this function times the derivative of the first; 3x² plus 5. That's your answer.

Your teacher may ask for you to simplify this answer. Assuming that they didn't want you to multiply this out in the beginning, they may not want you to multiply it out in the end. This is a very easy way to at least get a function, that is the derivative of your original polynomial.

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