Differentiation Rules

Differentiation rules are formulae that allow us to find the derivatives of functions quickly.

Taking derivatives of functions follows several basic rules:

• multiplication by a constant: $\big(c\cdot f(x)\big)' = c \cdot f'(x)$
• addition and subtraction: $\big( f(x)\pm g(x)\big)' = f'(x) \pm g'(x).$

For multiplication and composition of functions, see product rule and chain rule.

Main article: Power Rule

When taking the derivatives of polynomials, we can use the power rule:

Power Rule

$\frac{d}{dx} x^n = n\cdot x^{n-1}$

Main Article: Differentiation of Trigonometric Functions

We can see the basic trigonometric derivatives in the table below:

Main Article: Differentiation of Exponential Functions

The main formula you have to remember here is the derivative of a logarithm:

$\dfrac{d}{dx} \log_a x = \dfrac{1}{x \cdot \ln a}.$

What is the derivative of the following exponential function:

$f(x) = 2 \cdot 3^x ?$

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We have

$\begin{aligned} f'(x) &= 2(3^x)' \\ &= 2\cdot 3^x \ln 3\\ &= 2(\ln 3)\cdot 3^x. \ _\square \end{aligned}$

What is the derivative of the following exponential function:

$f(x) = 3e^x ?$

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We have

$\begin{aligned} f'(x) &= 3(e^x)' \\ &= 3e^x. \ _\square \end{aligned}$

Main Article: Differentiation of Logarithmic Functions

What is the derivative of the following logarithmic function:

$f(x) = \ln x + x ?$

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We have

$\begin{aligned} f'(x) &= (\ln x)' +x' \\ &= \frac{1}{x} + 1. \ _\square \end{aligned}$

What is the derivative of the following logarithmic function:

$f(x) =3\log_{2} x -x^2 +2 ?$

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We have

$\begin{aligned} f'(x) &= (3\log_{2}x)'-\big(x^2\big)' +0 \\ &= 3\cdot \frac{1}{x} \cdot \frac{1}{\ln 2} - 2x \\ &= \frac{3}{x\ln 2} - 2x. \ _\square \end{aligned}$

Main Article: Chain Rule

General form : $\frac{d}{dx} f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)$

What is the derivative of the following function:

$y=f(2x^2+x-3)?$

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We have

$\begin{aligned} y' &= f'(2x^2+x-3)\big(2x^2+x-3\big)' \\ &= (4x+1)f'(2x^2+x-3). \ _\square \end{aligned}$

What is the derivative of the following function:

$f(x) = (x^2+x+1)^5 ?$

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We have

$\begin{aligned} f'(x) &= 5\big(x^2+x+1\big)^4\big(x^2+x+1\big)' \\ &= 5\big(x^2+x+1\big)^4(2x+1). \ _\square \end{aligned}$

What is the derivative of the following function:

$y = \sqrt{2x+1} ?$

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Since $y = \sqrt{2x+1} = (2x+1)^{\frac{1}{2}} ,$ we have

$\begin{aligned} y' &= \frac{1}{2}(2x+1)^{\frac{1}{2}-1} \cdot (2x+1)' \\ &= (2x+1)^{-\frac{1}{2}} \\ &= \frac{1}{\sqrt{2x+1}}. \ _\square \end{aligned}$

Main Article: Product Rule

What is the derivative of the following function:

$y = (x^2+3)(x^3+2x^2+5) ?$

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We have

$\begin{aligned} y' &= (x^2+3)'(x^3+2x^2+5) + (x^2+3)(x^3+2x^2+5)' \\ &= 2x(x^3+2x^2+5) + (x^2+3)(3x^2+4x) \\ &= 5x^4 + 8x^3 + 9x^2 +22x. \ _\square \end{aligned}$

What is the derivative of the following function:

$y=(x^2+1)(2x+1)(3x+1)$

when $x = 1?$

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We have

$\begin{aligned} y' &= (x^2+1)'(2x+1)(3x+1) + (x^2+1)(2x+1)'(3x+1) + (x^2+1)(2x+1)(3x+1)' \\ &= 2x(2x+1)(3x+1) + (x^2+1)\cdot 2 \cdot (3x+1) + (x^2+1)(2x+1) \cdot 3 \\ &= 24x^3 + 15x^2 + 14x +5. \end{aligned}$

Thus, the derivative of $y$ at $x=1$ is

$y' \big\vert_{x=1} =24 \cdot 1^3 + 15 \cdot 1^2 + 14 \cdot 1 + 5 = 58. \ _\square$

Main Article: Quotient Rule

What is the derivative of the following function:

$y = \frac{1-x}{x^2+2} ?$

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We have

$\begin{aligned} y' &= \frac{(1-x)'(x^2+2)-(1-x)(x^2+2)'}{(x^2+2)^2} \\ &= \frac{(-1) \cdot (x^2+2)-(1-x) \cdot 2x}{(x^2+2)^2} \\ &= \frac{x^2-2x-2}{(x^2+2)^2}. \ _\square \end{aligned}$

What is the derivative of the following function:

$y = \frac{1}{x^2\sqrt{x^3}} ?$

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Since $y= \frac{1}{x^2\sqrt{x^3}} = x^{-\frac{7}{2}},$ we have $\begin{aligned} y' &= -\frac{7}{2} x^{-\frac{9}{2}} \\ &= -\frac{7}{2x^4\sqrt{x}}. \ _\square \end{aligned}$

Main Article: Differentiation of Inverse Functions

If $f(x) = x^3+ 7x+2,$ find $(f^{-1})'(10).$

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If $f$ is a one-to-one differentiable function with inverse function $f^{-1}$ and $f'\big(f^{-1}(a)\big) \neq 0,$ then the inverse function is differentiable at $a$ and

$(f^{-1})'(a)=\frac{1}{f'\big(f^{-1}(a)\big)}.$

Notice that $f$ of this problem is one-to-one because

$f'(x)=3x^2+7 > 0,$

so $f$ is increasing. To use the above theorem, we need to know $f^{-1}(10)$ and we can find it by equating $f(x)=10:$

$\begin{aligned} 10 &= x^3 + 7x + 2 \\ \Rightarrow x &= 1, f^{-1}(10) = 1. \end{aligned}$

Therefore,

$(f^{-1})'(10) = \frac{1}{f'\big(f^{-1}(10)\big)} = \frac{1}{f'(1)} = \frac{1}{3\cdot 1^2 + 7} = \frac{1}{10}. \ _\square$

If $f(x+y) = f(x) + f(y) + xy$ and $f'(0)=2,$ what is $f'(2)?$

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Substituting $x=0$ and $y=0$ into $f(x+y) = f(x) + f(y) + xy ,$ we get the value of $f(0)$ as follows:

$f(0) = f(0) + f(0) + 0 \implies f(0) = 0 .$

Thus,

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

If $f(x) = \left(x + \sqrt{1+x^2}\right)^{10},$ what is $f'(1)\cdot f'(-1) ?$

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We have

$\begin{aligned} f'(x) &= 10\left(x+\sqrt{1+x^2}\right)^9 \times \left(x+\sqrt{1+x^2}\right)' \\ &= 10\left(x+\sqrt{1+x^2}\right)^9\left(1 + \frac{2x}{2\sqrt{1+x^2}} \right ) \\ &= \frac{10}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^{10}. \end{aligned}$

Therefore,

$\begin{aligned} f'(1) \cdot f'(-1) &= \frac{10}{\sqrt{2}} \left(1+\sqrt{2}\right)^{10} \times \frac{10}{\sqrt{2}}\left(-1 + \sqrt{2}\right)^{10} \\ &= 50\left(\big(\sqrt{2} +1\big)\big(\sqrt{2} -1\big)\right)^{10} \\ &= 50(2-1)^{10} \\ &= 50. \ _\square \end{aligned}$

If $F(t) = f\big(g(t)\big), f'(t) = \frac{1}{t^2+1}, g'(t) = \frac{10}{t^4 + 1},$ and $g(0) = 3,$ what is $F'(0) ?$

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Since $F'(t) = f'\big(g(t)\big)g'(t),$

$\begin{aligned} G'(0) &= f'\big(g(0)\big)g'(0) \\\\ &= f'(3)\cdot \frac{10}{0^4+1} \\\\ &= \frac{1}{3^2 +1}\cdot 10 \\\\ &= 1. \ _\square \end{aligned}$

What is the derivatives of following function:

$y= \sqrt[3]{(x+1)(x^2+1)}$ when $x=0 ?$

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We have

$\begin{aligned} y &= \sqrt[3]{(x+1)(x^2+1)} \\ \Rightarrow y^3 &= (x+1)(x^2+1). \end{aligned}$

Differentiating both sides with respect to $x,$ we have

$\begin{aligned} 3y^2 \frac{dy}{dx} &= (x^2+1) +(x+1) \cdot 2x \\ &= 3x^2 +2x + 1 \\\\ \Rightarrow \frac{dy}{dx} &= \frac{3x^2 + 2x + 1}{3y^2} \\ &= \frac{3x^2+2x+1}{3 \sqrt[3]{(x+1)^2(x^2+1)^2}} \\\\ \Rightarrow \left. \frac{dy}{dx} \right\lvert_{x=0} &= \frac{1}{3}.\ _\square \end{aligned}$