State the constant, consistent multiple, and also power rules. Apply the amount and also distinction rules to integrate derivatives. Use the product dominion for finding the derivative of a product of attributes. Use the quotient preeminence for finding the derivative of a quotient of functions. Extfinish the power rule to functions via negative exponents. Combine the differentiation rules to discover the derivative of a polynomial or rational attribute.

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Finding derivatives of attributes by making use of the definition of the derivative deserve to be a prolonged and, for certain functions, a rather complicated process. For instance, formerly we uncovered that

by utilizing a procedure that associated multiplying an expression by a conjugate before evaluating a limit.

The procedure that we can usage to evaluate (dfracddxleft(sqrt<3>x ight)) utilizing the definition, while comparable, is even more complicated.

In this section, we build rules for finding derivatives that allow us to bypass this procedure. We start via the basics.



The Constant Rule

We initially use the limit meaning of the derivative to find the derivative of the constant function, (f(x)=c). For this function, both (f(x)=c) and (f(x+h)=c), so we achieve the complying with result:

<eginalign* f′(x) &=lim_h→0 dfracf(x+h)−f(x)h \<4pt> &=lim_h→0dfracc−ch \<4pt> &=lim_h→0dfrac0h \<4pt> &=lim_h→00=0. endalign*>

The dominance for separating continuous attributes is called the consistent rule. It claims that the derivative of a continuous attribute is zero; that is, considering that a continuous function is a horizontal line, the slope, or the price of readjust, of a constant feature is (0). We restate this dominion in the adhering to theorem.





The Power Rule

We have presented that

At this allude, you could see a pattern beginning to construct for derivatives of the form (dfracddxleft(x^n ight)). We continue our examination of derivative formulas by distinguishing power attributes of the develop (f(x)=x^n) wright here (n) is a positive integer. We construct formulas for derivatives of this kind of function in stperiods, start with positive integer powers. Before stating and proving the basic rule for derivatives of attributes of this form, we take a look at a certain instance, (dfracddx(x^3)). As we go through this derivation, pay distinct attention to the portion of the expression in boldchallenge, as the strategy provided in this situation is essentially the exact same as the technique provided to prove the general instance.



Exercise (PageIndex2)

Find (dfracddxleft(x^4 ight).)

Hint

Use ((x+h)^4=x^4+4x^3h+6x^2h^2+4xh^3+h^4) and also follow the procedure outlined in the coming before instance.

Answer

(dfracddxleft(x^4 ight) = 4x^3)


As we shall view, the procedure for finding the derivative of the general develop (f(x)=x^n) is exceptionally comparable. Although it is often unwise to draw general conclusions from specific examples, we note that when we identify (f(x)=x^3), the power on (x) becomes the coeffective of (x^2) in the derivative and the power on (x) in the derivative decreases by 1. The following theorem states that the power rule holds for all positive integer powers of (x). We will certainly ultimately extfinish this outcome to negative integer powers. Later, we will certainly watch that this rule might likewise be extfinished initially to rational powers of (x) and then to arbitrary powers of (x). Be mindful, however, that this preeminence does not use to features in which a consistent is elevated to a variable power, such as (f(x)=3^x).






The Sum, Difference, and also Constant Multiple Rules

We find our next differentiation rules by looking at derivatives of sums, distinctions, and constant multiples of features. Just as once we work-related with functions, tright here are rules that make it less complicated to find derivatives of features that we include, subtract, or multiply by a consistent. These rules are summarized in the complying with theorem.




Example (PageIndex4): Applying the Constant Multiple Rule

Find the derivative of (g(x)=3x^2) and compare it to the derivative of (f(x)=x^2.)

Solution

We use the power dominance directly:

Since (f(x)=x^2) has derivative (f′(x)=2x), we check out that the derivative of (g(x)) is 3 times the derivative of (f(x)). This connection is depicted in Figure.

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Figure (PageIndex1): The derivative of (g(x)) is 3 times the derivative of (f(x)).

Example (PageIndex5): Applying Basic Derivative Rules

Find the derivative of (f(x)=2x^5+7).

Solution

We start by applying the dominion for distinguishing the amount of two attributes, complied with by the rules for separating continuous multiples of attributes and also the dominion for separating powers. To much better understand the sequence in which the differentiation rules are used, we usage Leibniz notation throughout the solution:

(eginalign* f′(x)&=dfracddxleft(2x^5+7 ight)\<4pt>&=dfracddx(2x^5)+dfracddx(7) & & extApply the sum ascendancy.\<4pt>&=2dfracddx(x^5)+dfracddx(7) & & extApply the constant multiple rule.\<4pt>&=2(5x^4)+0 & & extApply the power dominance and also the constant dominance.\<4pt>&=10x^4 & & Simplify. endalign*)


Exercise (PageIndex4)

Find the derivative of (f(x)=2x^3−6x^2+3.)

Hint

Use the preceding example as a guide.

Answer

(f′(x)=6x^2−12x.)


Example (PageIndex6): Finding the Equation of a Tangent Line

Find the equation of the line tangent to the graph of (f(x)=x^2−4x+6) at (x=1)

Solution

To uncover the equation of the tangent line, we need a allude and a slope. To discover the point, compute

This provides us the allude ((1,3)). Since the slope of the tangent line at 1 is (f′(1)), we should initially find (f′(x)). Using the definition of a derivative, we have

so the slope of the tangent line is (f′(1)=−2). Using the point-slope formula, we check out that the equation of the tangent line is

Putting the equation of the line in slope-intercept develop, we obtain




Product Rule

Let (f(x)) and also (g(x)) be differentiable attributes. Then

That is,

< extif p(x)=f(x)g(x),quad extthen p′(x)=f′(x)g(x)+g′(x)f(x). onumber>

This means that the derivative of a product of 2 attributes is the derivative of the initially attribute times the second feature plus the derivative of the second attribute times the initially attribute.


Proof

We start by assuming that (f(x)) and (g(x)) are differentiable functions. At a vital suggest in this proof we should use the reality that, because (g(x)) is differentiable, it is likewise constant. In specific, we use the reality that given that (g(x)) is continuous, (displaystyle lim_h→0g(x+h)=g(x).)

By using the limit definition of the derivative to (p(x)=f(x)g(x),) we obtain

< p′(x)=lim_h→0dfracf(x+h)g(x+h)−f(x)g(x)h. onumber>

By including and also subtracting (f(x)g(x+h)) in the numerator, we have

After breaking apart this quotient and using the sum legislation for limits, the derivative becomes

Rearvarying, we obtain

<eginalign*p′(x)&=lim_h→0left(dfracf(x+h)−f(x)h⋅g(x+h) ight)+lim_h→0left(dfracg(x+h)−g(x)h⋅f(x) ight)\<4pt>&= left(lim_h→0dfracf(x+h)−f(x)h ight)⋅left(lim_h→0;g(x+h) ight)+left(lim_h→0dfracg(x+h)−g(x)h ight)⋅f(x)endalign*>

By using the continuity of (g(x)), the interpretation of the derivatives of (f(x)) and (g(x)), and applying the limit laws, we arrive at the product dominion,


Example (PageIndex7): Applying the Product Rule to Constant Functions

For (p(x)=f(x)g(x)), usage the product dominance to discover (p′(2)) if (f(2)=3,; f′(2)=−4,; g(2)=1), and also (g′(2)=6).

Solution

Since (p(x)=f(x)g(x)), (p′(x)=f′(x)g(x)+g′(x)f(x),) and hence

(p′(2)=f′(2)g(2)+g′(2)f(2)=(−4)(1)+(6)(3)=14.)


Example (PageIndex8): Applying the Product Rule to Binomials

For (p(x)=(x^2+2)(3x^3−5x),) uncover (p′(x)) by applying the product rule. Check the result by initially finding the product and then distinguishing.

Solution

If we set (f(x)=x^2+2) and (g(x)=3x^3−5x), then (f′(x)=2x) and (g′(x)=9x^2−5). Thus,

(p′(x)=f′(x)g(x)+g′(x)f(x)=(2x)(3x^3−5x)+(9x^2−5)(x^2+2).)

Simplifying, we have

To examine, we check out that (p(x)=3x^5+x^3−10x) and, in turn, (p′(x)=15x^4+3x^2−10.)




The Quotient Rule

Having emerged and also exercised the product rule, we currently take into consideration separating quotients of features. As we view in the following theorem, the derivative of the quotient is not the quotient of the derivatives; quite, it is the derivative of the feature in the numerator times the attribute in the denominator minus the derivative of the attribute in the denominator times the function in the numerator, all split by the square of the attribute in the denominator. In order to much better grasp why we cannot sindicate take the quotient of the derivatives, save in mind that


The Quotient Rule

Let (f(x)) and also (g(x)) be differentiable functions. Then

That is, if

then


Example (PageIndex9): Applying the Quotient Rule

Use the quotient dominion to discover the derivative of (q(x)=dfrac5x^24x+3.)

Solution

Let (f(x)=5x^2) and (g(x)=4x+3). Therefore, (f′(x)=10x) and (g′(x)=4).

Substituting into the quotient rule, we have

Simplifying, we obtain



Extended Power Rule

If (k) is an adverse integer, then


Proof

If (k) is an unfavorable integer, we might collection (n=−k), so that n is a positive integer through (k=−n). Due to the fact that for each positive integer (n),(x^−n=dfrac1x^n), we might now use the quotient dominance by establishing (f(x)=1) and also (g(x)=x^n). In this situation, (f′(x)=0) and (g′(x)=nx^n−1). Therefore,

Simplifying, we check out that

<eginalign* dfracdd(x^−n) &=dfrac−nx^n−1x^2n\<4pt>&=−nx^(n−1)−2n\<4pt>&=−nx^−n−1.endalign*>

Finally, observe that since (k=−n), by substituting we have


Example (PageIndex10): Using the Extfinished Power Rule

Find (dfracddx(x^−4)).

Solution

By using the extended power rule with (k=−4), we obtain


Example (PageIndex11): Using the Extfinished Power Rule and also the Constant Multiple Rule

Use the extfinished power dominance and the consistent multiple dominion to uncover (f(x)=dfrac6x^2).

Solution

It may seem tempting to usage the quotient rule to discover this derivative, and also it would definitely not be incorrect to execute so. However, it is much much easier to differentiate this feature by first rewriting it as (f(x)=6x^−2).

(eginalign* f′(x)&=dfracddxleft(dfrac6x^2 ight)=dfracddxleft(6x^−2 ight) & & extRecreate dfrac6x^2 ext as 6x^−2.\<4pt>&=6dfracddxleft(x^−2 ight) & & extApply the constant multiple preeminence.\<4pt>&=6(−2x^−3) & & extUse the extended power preeminence to differentiate x^−2.\<4pt>&=−12x^−3 & & extSimplify. endalign* )




Example (PageIndex12): Combining Differentiation Rules

For (k(x)=3h(x)+x^2g(x)), find (k′(x)).

Solution: Finding this derivative calls for the amount preeminence, the constant multiple dominion, and also the product ascendancy.

(k′(x)=dfracddxig(3h(x)+x^2g(x)ig)=dfracddxig(3h(x)ig)+dfracddxig(x^2g(x)ig)) Apply the amount preeminence.
(=3dfracddxig(h(x)ig)+left(dfracddx(x^2)g(x)+dfracddx(g(x))x^2 ight)) Apply the continuous multiple ascendancy to distinguish (3h(x)) and also the product dominion to identify (x^2g(x)).
(=3h′(x)+2xg(x)+g′(x)x^2)

Example (PageIndex13): Extending the Product Rule

For (k(x)=f(x)g(x)h(x)), express (k′(x)) in regards to (f(x),g(x),h(x)), and also their derivatives.

Solution

We can think of the function (k(x)) as the product of the function (f(x)g(x)) and the feature (h(x)). That is, (k(x)=(f(x)g(x))⋅h(x)). Hence,

(eginalign* k′(x)&=dfracddxig(f(x)g(x)ig)⋅h(x)+dfracddxig(h(x)ig)⋅ig(f(x)g(x)ig). & & extApply the product ascendancy to the product of f(x)g(x) ext and also h(x).\<4pt>&=ig(f′(x)g(x)+g′(x)f(x)ig)h(x)+h′(x)f(x)g(x) & & extApply the product dominion to f(x)g(x)\<4pt>&=f′(x)g(x)h(x)+f(x)g′(x)h(x)+f(x)g(x)h′(x). & & extSimplify.endalign* )


Example (PageIndex14): Combining the Quotient Rule and the Product Rule

For (h(x)=dfrac2x^3k(x)3x+2), uncover (h′(x)).

Solution

This procedure is typical for finding the derivative of a rational function.

(eginalign* h′(x)&=dfracdfracddx(2x^3k(x))⋅(3x+2)−dfracddx(3x+2)⋅(2x^3k(x))(3x+2)^2 & & extApply the quotient dominion.\<4pt>&=dfrac(6x^2k(x)+k′(x)⋅2x^3)(3x+2)−3(2x^3k(x))(3x+2)^2 & & extApply the product ascendancy to discover dfracddx(2x^3k(x)). ext Use dfracddx(3x+2)=3.\<4pt>&=dfrac−6x^3k(x)+18x^3k(x)+12x^2k(x)+6x^4k′(x)+4x^3k′(x)(3x+2)^2 & & extSimplify endalign* )


Exercise (PageIndex9)

Find (dfracddx(3f(x)−2g(x)).)

Hint

Apply the distinction preeminence and the continuous multiple ascendancy.

Answer

(3f′(x)−2g′(x).)


Example (PageIndex15): Determining Wright here a Function Has a Horizontal Tangent

Determine the worths of (x) for which (f(x)=x^3−7x^2+8x+1) has actually a horizontal tangent line.

Solution

To uncover the values of (x) for which (f(x)) has actually a horizontal tangent line, we must solve (f′(x)=0).

Since (f′(x)=3x^2−14x+8=(3x−2)(x−4)),

we need to deal with ((3x−2)(x−4)=0). Therefore we see that the feature has horizontal tangent lines at (x=dfrac23) and also (x=4) as presented in the adhering to graph.

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api/deki/files/2279/CNX_Calc_Figure_03_03_003.jpeg?revision=1&size=bestfit&width=666&height=307" />Figure (PageIndex3): The grandstand also beside a straightameans of the Circuit de Barcelona-Catalunya race track, located wbelow the spectators are not in hazard.

Safety is especially a problem on turns. If a driver does not sluggish down enough prior to entering the revolve, the car might slide off the racetrack. Typically, this just results in a more comprehensive rotate, which slows the driver dvery own. But if the driver loses control entirely, the car might fly off the track totally, on a course tangent to the curve of the racetrack.

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Suppose you are making a brand-new Formula One track. One area of the track have the right to be modeled by the attribute (f(x)=x^3+3x^2+x) (Figure (PageIndex4)). The existing setup calls for grandstands to be built along the first straightamethod and also roughly a portion of the first curve. The plans call for the front corner of the grandstand also to be situated at the suggest ((−1.9,2.8)). We desire to identify whether this place puts the spectators in peril if a driver loses manage of the auto.

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Figure (PageIndex4): (a) One area of the racetrack have the right to be modeled by the function (f(x)=x^3+3x^2+x). (b) The front edge of the grandstand also is situated at ((−1.9,2.8)). Physicists have figured out that drivers are a lot of most likely to shed regulate of their cars as they are coming right into a rotate, at the allude wright here the slope of the tangent line is 1. Find the ((x,y)) collaborates of this suggest near the revolve. Find the equation of the tangent line to the curve at this point. To determine whether the spectators are in danger in this scenario, uncover the (x)-coordinate of the allude where the tangent line crosses the line (y=2.8). Is this point safely to the ideal of the grandstand? Or are the spectators in danger? What if a driver loses regulate earlier than the physicists project? Suppose a driver loses control at the suggest ((−2.5,0.625)). What is the slope of the tangent line at this point? If a driver loses control as described in part 4, are the spectators safe? Should you proceed through the current architecture for the grandstand also, or need to the grandstands be moved?