Concept of the Derivative - AP Calculus AB

Card 0 of 957

Question

Evaluate the following difference quotient:

$\lim_{x\rightarrow \Delta x}\frac{cot(x+\Delta x)-cot(x)}{\Delta x}$

Answer

One should recognize the difference quotient as being equivalent to the derivative or instantaneous rate of change.

That is, for some continuous function, .

$f'(x)=\lim_{x\rightarrow \Delta x}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

The function in our question is

We note from this that because the difference quotient is the derivative:

$\lim_{x\rightarrow \Delta x}\frac{cot(x+\Delta x)-cot(x)}{\Delta x}=-csc^2(x)$

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Question

Find the derivative of the following function:

Answer

We use the power rule on each term of the function.

The first term

becomes

$2 \times 6 x^2 = 12 x$ .

The second term

becomes

.

The final term, 7, is a constant, so its derivative is simply zero.

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Question

What is the derivative of ?

Answer

To get $\frac{\mathrm{d} }{\mathrm{d} x}(2x)$ , we can use the power rule.

Since the exponent of the is , as , we lower the exponent by one and then multiply the coefficient by that original exponent:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=12x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2x^{0}$

Anything to the power is .

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=21$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2$

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2x^{2-1})+(15x^{1-0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5x^0$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5$

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as , as anything to the zero power is one.

That means this problem will look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})+(08x^{0-1})$

Notice that $(08x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{0})$

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=51$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=?$

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})+(08x^{0-1})$

Notice that $08x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

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Question

What is the derivative of ?

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})+(0x^{0-1})$

Notice that $(0x^{0-1})=0$ since anything times zero is zero.

That leaves us with $\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})$ .

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(x^{0})$

As stated earlier, anything to the zero power is one, leaving us with:

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=1$

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Question

What is the derivative of ?

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})+(04x^{0-1})$

Notice that $(04x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{1})+(113x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

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Question

What is the derivative of $2x^4+\frac{1}{2}x$ ?

Answer

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{4-1})+(1\frac{1}{2}x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

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Question

What is the derivative of ?

Answer

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

We are going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(012x^{0-1})$

Notice that $(012x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3$

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Question

What is the derivative of ?

Answer

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{2-1})+(112x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{1})+(112x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

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Question

What is the derivative of ?

Answer

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

For this problem that would look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{2-1})+(015x^{0-1})$

Notice that $(015x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

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Question

What is the derivative of ?

Answer

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{3-1})+(22x^{2-1})+(05x^{0-1})$

Notice that $(05x^{0-1}) =0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{3-1})+(22x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{2})+(22x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=3x^2+4x$

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Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left {(x+h)^2 +(x+h) \right } -\sin\left {h^2+h \right }}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Answer

$L(x)=\lim_{h->0}\frac{\sin\left {(x+h)^2 +(x+h) \right } -\sin\left {h^2+h \right }}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left {(x+h)^2 +(x+h) \right } -\sin\left {h^2+h \right }}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left { (x+h)^2+(x+h) \right }$

$f(h) = \sin\left { h^2+h\right }$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

_

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left { (x+h)^2+(x+h) \right }+C$

$f(h) = \sin\left { h^2+h\right }+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

_

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left {(x+h)^2 +(x+h) \right } -\sin\left {h^2+h \right }}{h} =(2x+1)\cos(x^2+x)$

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Question

Evaluate $\lim_{h \to 0} \frac{5(2+h)^{10}-5(2)^{10}}{h}$ .

Answer

This limit can't be evaluated by conventional limit laws. To see why the answer is , we have to recognize that the limit looks like

$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ , with $f(x)=5x^{10}, x=2$ .

This new limit is a conventional expression for , with substituted in for

We can find with the power rule, and substituting gives .

To summarize, since

$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=f'(x)$ , we have

$\lim_{h \to 0} \frac{5(2+h)^{10}-5(2)^{10}}{h} = f'(2)=50(2)^9 = 25600$ .

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Question

Calculate the limit by interpreting it as the definition of the derivative of a function.

$\lim_{x->a}\frac{\frac{1}{(x-a)^2}-\frac{1}{a^2}}{x-a}$

Answer

$\lim_{x->a}\frac{\frac{1}{(x-a)^2}-\frac{1}{a^2}}{x-a}$

There are two commonly used formulations of the difference quotient used to compute derviatives. Both definitions are equivalent and should always give the same derivative for a given function .

$\lim_{x->a}\frac{f(x)-f(a)}{x-a}$

$\lim_{h->0}\frac{f(x+h)-f(x)}{h}$

The limit in this particular problem resembles the first difference quotient listed above.

$\lim_{x->a}\frac{f(x)-f(a)}{x-a}=\lim_{x->a}\frac{\frac{1}{(x-a)^2}-\frac{1}{a^2}}{x-a}$

By inspection, it's clear that the in our case must be $\frac{1}{x^2}$ . We are being asked to evaluate the limit by interpreting it as the definition of the derivative of a function. We must therefore identify the function, which is clearly $\frac{1}{x^2}$ , and then differentiate it using known rules of differentiation.

First rewrite the function to apply the rule for $\frac{d}{dx}nx^{n-1}$ :

$f(x)=\frac{1}{x^2}=x^{-2}$

$f'(x)=-2x^{-3}=-\frac{2}{x^3}$

This means

$\lim_{x->a}\frac{\frac{1}{(x-a)^2}-\frac{1}{a^2}}{x-a}=-\frac{2}{x^3}$

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Question

Give the difference quotient of the function

$f(x) = e^{2x}$

Answer

The difference quotient of a function is the expression

$\frac{f(x+h) - f(x)}{h}$ .

If $f(x) = e^{2x }$ , this expression is

$\frac{e^{2(x+h)}-e^{2x}}{h}$

$=\frac{e^{2x+2h}-e^{2x}}{h}$

$=\frac{e ^{2h} e^{2x }-1e^{2x}}{h}$

$=\left (\frac{e ^{2h} -1 }{h} \right )e^{2x }$

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Question

Find the derivative of the following function:

Answer

The derivative of a function can be defined according to the following:

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Using this for our function - and not forgetting to write the limit for every step! - we get

$\lim_{h\rightarrow 0} \frac{5(x+h)+2-(5(x)+2)}{h}= \lim_{h\rightarrow 0} \frac{5x+5h+2-5x-2}{h}=\lim_{h\rightarrow 0} 5=5$

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Question

Find the derivative of the function using the limit of the difference quotient:

Answer

The derivative of a function as defined by the limit of the difference quotient is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Evaluating the limit using our function - and always writing the limit symbol! - we get

$\lim_{h\rightarrow 0}\frac{2(x+h)^2+3-(2x^2+3)}{h}= \lim_{h\rightarrow 0}\frac{2(x^2+2xh+h^2)+3-2x^2-3}{h}=\lim_{h\rightarrow 0}\frac{4xh+2h^2}{h}=\lim_{h\rightarrow0}4x+2h=4x$

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Question

Find the derivative of the function using the limit of the difference quotient

Answer

To find the derivative using the limit of the difference quotient, we use the formula:

$f'=\lim_{h\rightarrow 0}\left [ \frac{f(x+h)-f(x)}{h}\right ]$

$f'=\lim_{h\rightarrow 0}\left [ \frac{(x^2+2xh+h^2)+2(x+h)+1-(x^2+2x+1) }{h}\right ]$

$f'=\lim_{h\rightarrow 0}\frac{2xh+h^2+2h}{h}=2x+2$

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