3.1 Product Ruleap Calculus

3.1 Product Ruleap Calculus Solver
3.1 Product Ruleap Calculus Formulas
3.1 Product Ruleap Calculus 14th Edition
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F of X equals X to the 4th and G of X equals X to the 6th. So I can make a product out of the two functions. F times G and that'll turn out to be X to the 10th. Now what's the derivative of this product of the two functions? Well you might expect the derivative to be, the derivative of the product to be the product of the derivatives. Essential Calculus. Franklin Wright, Spencer P. Hurd, and Bill D. Essential Calculus introduces students to basic concepts in the field of calculus. Each chapter section provides examples including graphs, tables, and diagrams. Precalculus & Elements of Calculus tutorial videos. One Bernard Baruch Way (55 Lexington Ave. At 24th St) New York, NY 10010 646-312-1000.

We start with the derivative of a power function, $ds f(x)=x^n$. Here $n$ is a number of any kind: integer,rational, positive, negative, even irrational, as in $ds x^pi$. We havealready computed some simple examples, so the formula should not be acomplete surprise:$${dover dx}x^n = nx^{n-1}.$$It is not easy to show this is true for any $n$. We will do some ofthe easier cases now, and discuss the rest later.

The easiest, and most common, is the case that $n$ is a positiveinteger. To compute the derivative we need to compute the followinglimit:$${dover dx}x^n = lim_{Delta xto0} {(x+Delta x)^n-x^nover Delta x}.$$For a specific, fairly small value of $n$, we could do this bystraightforward algebra.

3.1 Product Ruleap Calculus Solver

Example 3.1.1 Find the derivative of $ds f(x)=x^3$.$$eqalign{{dover dx}x^3 &= lim_{Delta xto0} {(x+Delta x)^3-x^3over Delta x}.cr&=lim_{Delta xto0} {x^3+3x^2Delta x+3xDelta x^2 + Delta x^3-x^3over Delta x}.cr&=lim_{Delta xto0}{3x^2Delta x+3xDelta x^2 + Delta x^3over Delta x}.cr&=lim_{Delta xto0}3x^2+3xDelta x + Delta x^2 = 3x^2.cr}$$

The general case is really not much harder as long as we don't try todo too much. The key is understanding what happens when $ds (x+Delta x)^n$is multiplied out:$$(x+Delta x)^n=x^n + nx^{n-1}Delta x + a_2x^{n-2}Delta x^2+cdots++a_{n-1}xDelta x^{n-1} + Delta x^n.$$We know that multiplying out will give a large number of terms all ofthe form $ds x^iDelta x^j$, and in fact that $i+j=n$ in every term. Oneway to see this is to understand that one method for multiplying out $ds (x+Delta x)^n$ is the following: In every $(x+Delta x)$ factor,pick either the $x$ or the $Delta x$, then multiply the $n$ choicestogether; do this in all possible ways. For example, for $ds (x+Delta x)^3$, there are eight possible ways to do this:$$eqalign{(x+Delta x)(x+Delta x)(x+Delta x)&=xxx + xxDelta x + xDelta x x+ xDelta x Delta xcr&qquad+ Delta x xx + Delta xxDelta x + Delta xDelta x x+ Delta xDelta x Delta xcr&= x^3 + x^2Delta x +x^2Delta x +xDelta x^2cr&quad+x^2Delta x +xDelta x^2 +xDelta x^2 +Delta x^3cr&=x^3 + 3x^2Delta x + 3xDelta x^2+Delta x^3cr}$$No matter what $n$ is, there are $n$ ways to pick $Delta x$ in onefactor and $x$ in the remaining $n-1$ factors; this means one term is$ds nx^{n-1}Delta x$. The other coefficients are somewhat harder tounderstand, but we don't really need them, so in the formula abovethey have simply been called $ds a_2$, $ds a_3$, and so on. We know that everyone of these terms contains $Delta x$ to at least the power 2. Nowlet's look at the limit:$$eqalign{{dover dx}x^n &= lim_{Delta xto0} {(x+Delta x)^n-x^nover Delta x}cr&=lim_{Delta xto0} {x^n + nx^{n-1}Delta x + a_2x^{n-2}Delta x^2+cdots+a_{n-1}xDelta x^{n-1} + Delta x^n-x^nover Delta x}cr&=lim_{Delta xto0} {nx^{n-1}Delta x + a_2x^{n-2}Delta x^2+cdots+a_{n-1}xDelta x^{n-1} + Delta x^nover Delta x}cr&=lim_{Delta xto0} nx^{n-1} + a_2x^{n-2}Delta x+cdots+a_{n-1}xDelta x^{n-2} + Delta x^{n-1} = nx^{n-1}.cr}$$

3.1 Product Ruleap Calculus Formulas

Now without much trouble we can verify the formula for negativeintegers. First let's look at an example:

Example 3.1.2 Find the derivative of $ds y=x^{-3}$. Using the formula,$ds y'=-3x^{-3-1}=-3x^{-4}$.

Here is the general computation. Suppose $n$ is a negative integer;the algebra is easier to follow if we use $n=-m$ in the computation,where $m$ is a positive integer.$$eqalign{{dover dx}x^n &= {dover dx}x^{-m} =lim_{Delta xto0} {(x+Delta x)^{-m}-x^{-m}over Delta x}cr&=lim_{Delta xto0} { {1over (x+Delta x)^m} - {1over x^m} over Delta x} cr&=lim_{Delta xto0} { x^m - (x+Delta x)^m over(x+Delta x)^m x^m Delta x} cr&=lim_{Delta xto0} { x^m - (x^m + mx^{m-1}Delta x + a_2x^{m-2}Delta x^2+cdots+a_{m-1}xDelta x^{m-1} + Delta x^m)over(x+Delta x)^m x^m Delta x} cr&=lim_{Delta xto0} { -mx^{m-1} - a_2x^{m-2}Delta x-cdots-a_{m-1}xDelta x^{m-2} - Delta x^{m-1})over(x+Delta x)^m x^m} cr&={ -mx^{m-1} over x^mx^m}={ -mx^{m-1} over x^{2m}}=-mx^{m-1-2m}= nx^{-m-1} = nx^{n-1}.cr}$$

We will later see why the other cases of the power rule work, but fromnow on we will use the power rule whenever $n$ is any real number.Let's note here a simple case in which the power rule applies, oralmost applies, but is not really needed. Suppose that $f(x)=1$;remember that this '1' is a function, not 'merely' a number, andthat $f(x)=1$ has a graph that is a horizontal line, with slope zeroeverywhere. So we know that $f'(x)=0$. We might also write $ds f(x)=x^0$,though there is some question about just what this means at $x=0$. Ifwe apply the power rule, we get $ds f'(x)=0x^{-1}=0/x=0$, again notingthat there is a problem at $x=0$. So the power rule 'works' in thiscase, but it's really best to just remember that the derivative of anyconstant function is zero.

Exercises 3.1

Find the derivatives of the given functions.

Ex 3.1.1$ds x^{100}$(answer)

Ex 3.1.2$ds x^{-100}$(answer)

Ex 3.1.3$displaystyle {1over x^5}$(answer)

Ex 3.1.4$ds x^pi$(answer)

Ex 3.1.5$ds x^{3/4}$(answer)

Ex 3.1.6$ds x^{-9/7}$(answer)