Derivatives of Polynomials

Polynomials are one of the simplest functions to differentiate. When taking derivatives of polynomials, we primarily make use of the power rule.

Power Rule

For a real number $n$, the derivative of $f(x)= x^n$ is

$$\frac{d}{dx} f(x) = n x ^{n-1}.$$

Given a linear function $f(x) = ax+b$, we use the property that differentiation is linear to show

$$\frac{d}{dx} f(x) = \frac{d}{dx} ( ax + b ) = \frac{d}{dx} ( ax ) + \frac{d}{dx} (b) = a.$$

Thus, the derivative of a linear function is a constant.

If $f(x) = 3x + 2$, what is $f'(x)?$

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We have $f'(x) = 3,$ the slope of the line. $_\square$

If $f(x)$ is a linear function such that $f(1) = 2$ and $f'(3) = 4$, what is $f(x)?$

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Let $f(x)$ be the linear function $f(x)= ax + b$. Since $f'(3) = 4$, we have $a = 4$. Since $f(1) = 2$, we have $2 = 4 \times 1 + b$, implying $b = - 2$.

Thus, $f(x) = 4x - 2$. $_\square$

If $f(x)$ is a linear function with $f(2)=10$ and $f'(1)=4,$ what is the value of

$$f(0)+f'(0)?$$

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Let $f(x)$ be the linear function $f(x) = ax + b$. Since $a=f'(1)=4$, we have $f(x) = 4x + b$. Then $10 = f(2) = 4(2) + b$, implying $b=2$. Therefore, $f(0) = 4(0) + 2 = 2$, $f'(0) = a = 4$, and

$$f(0) + f'(0) = 2 + 4 = 6. \ _\square$$

If $f(x) = 3x^2 + 4x + 5$, what is $f'(1)?$

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We have $f'(x) = 2 \times 3 x + 1 \times 4 = 6x + 4$, which implies $f'(1) = 10$. $_\square$

If $f(x) = -3x^4 + 7x^3 - 6x^2 - \pi x + 5$, what is $\frac{d}{dx} f(x)?$

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We have

$$\begin{aligned} \frac{d}{dx} \big( -3x^4 + 7x^3 - 6x^2 - \pi x + 5 \big) &= (-3)(4) x^{4-1} + (7)(3) x^{3-1} - (6)(2) x^{2-1} - (\pi) (1) x^{1-1} + 0\\ &= -12x^3 + 21x^2 - 12x -\pi. \ _\square \end{aligned}$$

If $f(x) = (3x-2) ^4$, what is $\frac{d}{dx} f(x)?$

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Note that $f(x)$ is a polynomial but is not in the form given in the summary above. We will later see methods to differentiate this function directly, but to use the tools we have so far, we first expand the polynomial by the binomial theorem. This gives

$$\begin{aligned} f(x) &= (3x-2)^4 \\ &= (3x)^4 + 4(3x)^3(-2) + \frac{4 \cdot 3}{2} (3x)^2 (-2)^2 + \frac{4\cdot 3 \cdot 2}{1 \cdot 2 \cdot 3} (3x)(-2)^3 + (-2)^4\\ &= 81 x^4 - 216 x^3 + 216 x^2 - 96 x + 16, \end{aligned}$$

which implies

$$\begin{aligned} \frac{d}{dx} f(x) &= (81)(4)x^3 - (216)(3)x^2 + (216)(2) x - 96 \\ & = 324 x^3 - 648x^2 + 432 x - 96. \ _\square \end{aligned}$$

If $f(x) = x^3$, what is ${\displaystyle \lim_ { h \rightarrow 0 }}\frac{ f(2+h) - f(2) } { h }?$

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Using the above formula, we have

$$\begin{aligned} \lim_ { h \rightarrow 0 } \frac{ f(2+h) - f(2) } { h } &= \lim_ { h \rightarrow 0 } \frac{ (2+h)^3 - 2^3 } { h }\\ &= 3 \times 2^{3-1}\\ &= 12. \ _\square \end{aligned}$$

If ${\displaystyle\lim_ { h \rightarrow 0 }} \frac{ f(x+h) - f(x) } { h } = 4x^2$, what is $f'(2)?$

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Using the above formula, we know ${\displaystyle \lim_ { h \rightarrow 0 }} \frac{ f(x+h) - f(x) } { h } = \frac{d}{dx}f(x) = f'(x) = 4x^2$. Then

$$\begin{aligned} f'(2) &=\lim_ { h \rightarrow 0 } \frac{ f(2+h) - f(2) } { h } \\ &= 4\times 2^2 \\ &= 16. \ _\square \end{aligned}$$

If $f(x) = x^2 + 4x$, what is ${\displaystyle\lim_ { h \rightarrow 3 }} \frac{ f(h) - f(3) } { h -3}?$

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Note that ${\displaystyle\lim_ { h \rightarrow 3 }} \frac{ f(h) - f(3) } { h -3}$ is not in the form given in the formula above. Thus, we use the substitution method in order to get a general formula. Substituting $k$ for $h - 3,$ we have

$$\lim_ { h \rightarrow 3 } \frac{ f(h) - f(3) } { h -3} = \lim_ { k \rightarrow 0 } \frac{ f(3+k) - f(3) } {k}.$$

Then since ${\displaystyle\lim_ { h \rightarrow 0 }} \frac{ f(x+h) - f(x) } { h } = \frac{d}{dx}f(x) = f'(x) = 2x + 4$, we have

$$\lim_ { k \rightarrow 0 } \frac{ f(3+k) - f(3) } {k}=f'(3) = 2 \times 3 + 4 = 10. \ _\square$$

Find the derivative of $f(x) = (ax + b)^n$ for $x > -\frac{b}{a}.$

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Write $u = ax + b \implies u' = a$ so that $f(x) = u^n.$

Then $f'(x) = \frac{d}{dx} (u^n) = nu^{n - 1} \cdot u' = an(ax + b)^{n - 1} .\ _\square$

The derivative of the function $x^n$, where $n$ is a non-zero real number, is $n x ^{n-1}$.

For a positive integer $n$, we can prove this by first principles, using the binomial theorem:

$$\begin{aligned} \lim_ { h \rightarrow 0 } \frac{ ( x+h)^n - x^n } { h } & = \lim_{ h \rightarrow 0 } \frac{ \left[ x^n + n x^{n-1} h + n(n-1) x^{n-2} h^2 + \cdots + h^n \right] - x^n } { h} \\ & = \lim_{h \rightarrow 0 } n x^{n-1} + n(n-1) x^{n-2} h + \cdots + h^{n-1}\\ & = n x^{n-1}. \end{aligned}$$

By linearity of differentiation we conclude that if $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_ 0$, then

$$\frac{ d}{dx} f(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-1} + \cdots + 2 a_2 x + 1 a_1 x^0.$$

If $f(x) = (x^2 - 1 ) ( x^2 - 2x + 3)$, what is $f'(1)?$

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Multiplying out, we get $f(x) = x^4 - 2x^3 + 2x^2 + 2x - 3$. Hence $f'(x) = 4x^3 - 6x^2 + 4x + 2$. Thus

$$f'(1) = 4 - 6 + 4 + 2 = 4. \ _\square$$

Note: We can also use the product rule to differentiate this function, to conclude that $f'(x) = (x^2 - 1 ) ( 2x - 2 ) + ( 2x) ( x^2 - 2x + 3 ).$

Find the derivative of $f(x) = (ax + b)^n(cx + d)^m.$

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We have

$$\begin{aligned} f(x) & = \underbrace{(ax + b)^n}_{u(x)} \cdot \underbrace{(cx + d)^m}_{v(x)} \\\\ f'(x) & = u'(x)v(x) + u(x)v'(x) \\ & = \underbrace{n(ax + b)^{n - 1}(a)}_{u'(x)}(cx + d)^m + (ax + b)^n\underbrace{m(cx + d)^{m - 1}(c)}_{v'(x)} \\ & = (ax + b)^n(cx + d)^m\big(na(ax + d)^{-1} + mc(ax + b)^{-1}\big) \\ & = (ax + b)^n(cx + d)^m \left(\dfrac{na}{ax + b} + \dfrac{mc}{cx + d}\right).\ _\square \\ \end{aligned}$$

Find the derivative of $f(x) = \sqrt{x} + 2x^\frac34 + 3x^\frac56.$

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We have

$$\begin{aligned} f(x) & = x^\frac12 + 2x^\frac34 + 3x^\frac56 \\\\ f'(x) & = \dfrac12 x^{\frac12 - 1} + 2\cdot \dfrac34 x^{\frac34 - 1} + 3 \cdot \dfrac56 x^{\frac56 - 1} \\ & = \dfrac{1}{2\sqrt{x}} + \dfrac{3}{2 \sqrt[4]{x}} + \dfrac{5}{2 \sqrt[6]{x}}.\ _\square \\ \end{aligned}$$