Differentiation

Answer: $4e^{x} - 4\ln(10) \cdot 10^{x}$

$e^{x}$ is its own derivative, and $10^{x}$ has the derivative $10^{x}\ln(10)\mathclose{.}$ So $f'(x) = 4e^{x} - 4\ln(10) \cdot 10^{x}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Treated $10^{x}$ like a power of $x\mathclose{,}$ as $x \cdot 10^{x - 1}$ → $4e^{x} - 4x \cdot 10^{x - 1}$
• Forgot the factor $\ln(10)$ → $4e^{x} - 4 \cdot 10^{x}$

Answer: $h + 2$

$\dfrac{f(3 + h) - f(3)}{h}\mathclose{.}$ Expand $f(3 + h) = h^{2} + 2 h - 2$ and subtract $f(3) = -2\mathclose{:}$ the difference is $h^{2} + 2 h\mathclose{,}$ and every term has a factor $h\mathclose{.}$ Divided by $h\mathclose{:}$ $h + 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not divide by $h$ → $h^{2} + 2 h$
• Expanded $(3 + h)^2$ as $9 + h^2$ → $h - 4$

Answer: $5x^{4}\left(\cos(5x) - x\sin(5x)\right)$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{5}$ and $v = \cos(5x)\mathclose{:}$ $u' = 5x^{4}$ and $v' = -5\sin(5x)\mathclose{.}$ So $f'(x) = 5x^{4}\cos(5x) + x^{5} \cdot \left(-5\sin(5x)\right)\mathclose{,}$ which factorises as $5x^{4}\left(\cos(5x) - x\sin(5x)\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $-25x^{4}\sin(5x)$
• Forgot the factor $5$ that the chain rule gives inside $\cos(5x)$ → $x^{4}\left(5\cos(5x) - x\sin(5x)\right)$
• Took the derivative of $\cos$ to be $+\sin$ → $x^{4}\left(5\cos(5x) + 5x\sin(5x)\right)$

Answer: $\left(8 x + 5\right)e^{4 x^{2} + 5 x}$

Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside: $e^{4 x^{2} + 5 x} \cdot \left(8 x + 5\right) = \left(8 x + 5\right)e^{4 x^{2} + 5 x}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to multiply by the derivative of the exponent → $e^{4 x^{2} + 5 x}$
• Brought the exponent down like a power, $u\,e^{u - 1}$ → $\left(4 x^{2} + 5 x\right)e^{4 x^{2} + 5 x - 1}$

Answer: $\dfrac{2}{x} + \dfrac{5}{x\ln(10)}$

$\ln(x)$ has the derivative $\dfrac{1}{x}\mathclose{.}$ And $\log_{10}(x) = \dfrac{\ln(x)}{\ln(10)}\mathclose{,}$ so its derivative is $\dfrac{1}{x\ln(10)}\mathclose{.}$ So $f'(x) = \dfrac{2}{x} + \dfrac{5}{x\ln(10)}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot the $\ln(10)$ in the derivative of $\log_{10}(x)$ → $\dfrac{7}{x}$
• Multiplied by $\ln(10)$ instead of dividing by it → $\dfrac{2}{x} + \dfrac{5\ln(10)}{x}$

Answer: $18$

The difference quotient is the change in $f$ divided by the change in $x\mathclose{:}$ $\dfrac{f(4) - f(0)}{4 - 0} = \dfrac{76 - 4}{4} = 18\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not divide by the length of the interval → $72$
• Subtracted the $x$-values the other way round → $-18$

Answer: $2x\left(x + 1\right)e^{2x}$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{2}$ and $v = e^{2x}\mathclose{:}$ $u' = 2x$ and $v' = 2e^{2x}\mathclose{.}$ So $f'(x) = 2xe^{2x} + x^{2} \cdot 2e^{2x}\mathclose{,}$ which factorises as $2x\left(x + 1\right)e^{2x}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $4xe^{2x}$
• Forgot the factor $2$ that the chain rule gives inside $e^{2x}$ → $x\left(x + 2\right)e^{2x}$

Answer: $20 x^{2} - 100 x + 133$

Put $g(x)$ in place of every $x$ in the outer function: $f(g(x)) = 5\left(2 x - 5\right)^{2} + 8\mathclose{,}$ and $\left(2 x - 5\right)^{2} = 4 x^{2} - 20 x + 25\mathclose{.}$ Altogether $f(g(x)) = 20 x^{2} - 100 x + 133\mathclose{.}$

Common mistakes, and the answer each one gives:

• Composed the wrong way round, working out $g(f(x))$ → $10 x^{2} + 11$
• Squared each term of $g(x)$ separately → $20 x^{2} + 133$
• Multiplied $f(x)$ by $g(x)$ instead of composing them → $10 x^{3} - 25 x^{2} + 16 x - 40$

Answer: $\dfrac{- 5 x^{2} + 8 x + 15}{\left(x^{2} + 3\right)^{2}}$

The quotient rule is $\dfrac{u'v - uv'}{v^2}$ — and the ORDER of the two products matters, unlike the product rule. With $u = 5 x - 4$ and $v = x^{2} + 3\mathclose{:}$ $u' = 5$ and $v' = 2x\mathclose{.}$ The numerator is $5\left(x^{2} + 3\right) - \left(5 x - 4\right) \cdot 2x = - 5 x^{2} + 8 x + 15\mathclose{,}$ so $f'(x) = \dfrac{- 5 x^{2} + 8 x + 15}{\left(x^{2} + 3\right)^{2}}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Wrote $uv' - u'v$ in the numerator, the two products swapped → $\dfrac{5 x^{2} - 8 x - 15}{\left(x^{2} + 3\right)^{2}}$
• Differentiated top and bottom separately → $\frac{5}{2 x}$

Answer: $10 x^{4} - 4 x - 4$

Term by term (the sum rule): $x^n$ becomes $nx^{n-1}\mathclose{,}$ a number in front stays in front, and the constant $5$ becomes $0\mathclose{.}$ So $f'(x) = 10 x^{4} - 4 x - 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Kept the constant $5$ instead of making it $0$ → $10 x^{4} - 4 x + 1$
• Multiplied by each power but did not lower it → $10 x^{5} - 4 x^{2} - 4 x$

Answer: $40x\left(5 x^{2} - 4\right)^{3}$

Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside: $4\left(5 x^{2} - 4\right)^{3} \cdot 10x = 40x\left(5 x^{2} - 4\right)^{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to multiply by the derivative of the inside → $4\left(5 x^{2} - 4\right)^{3}$
• Differentiated the inside and raised THAT to the power → $10000 x^{4}$

Answer: $\dfrac{3}{\sqrt{x}} + \dfrac{6}{x^{4}}$

Write both as powers: $f(x) = 6x^{\frac{1}{2}} - 2x^{-3}\mathclose{.}$ The power rule lowers each power by one: $f'(x) = 3x^{-\frac{1}{2}} + 6x^{-4} = \dfrac{3}{\sqrt{x}} + \dfrac{6}{x^{4}}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the $\frac{1}{2}$ that the power of the root brings → $\dfrac{6}{\sqrt{x}} + \dfrac{6}{x^{4}}$
• Lowered the power $-3$ towards zero, to $-2$ → $\dfrac{3}{\sqrt{x}} + \dfrac{6}{x^{2}}$

Answer: $6$

On $[3, 3 + h]$ the difference quotient simplifies to $2 h + 6\mathclose{.}$ As $h$ goes to $0$ that closes in on $6\mathclose{:}$ the slope at $x = 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took $h = 1$ instead of letting $h$ go to $0$ → $8$
• Gave the function value $f(3)$ instead → $4$

Answer: $x^{3}\left(4\ln(x) + 1\right)$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{4}$ and $v = \ln(x)\mathclose{:}$ $u' = 4x^{3}$ and $v' = \dfrac{1}{x}\mathclose{.}$ So $f'(x) = 4x^{3}\ln(x) + x^{4} \cdot \dfrac{1}{x}\mathclose{,}$ which factorises as $x^{3}\left(4\ln(x) + 1\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $4x^{2}$
• Differentiated only $x^{4}$ and kept $\ln(x)$ as it was → $4x^{3}\ln(x)$

Answer: $\dfrac{5}{\cos^2(x)}$

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}\mathclose{,}$ and the quotient rule gives $\dfrac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \dfrac{1}{\cos^2(x)}\mathclose{.}$ So $f'(x) = \dfrac{5}{\cos^2(x)}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided the derivatives of the sine and the cosine → $-\dfrac{5\cos(x)}{\sin(x)}$
• Gave $\dfrac{1}{\sin^2(x)}$ in place of $\dfrac{1}{\cos^2(x)}$ → $\dfrac{5}{\sin^2(x)}$

Answer: $\dfrac{5}{2\sqrt{5 x - 8}}$

Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside: $\dfrac{1}{2\sqrt{5 x - 8}} \cdot 5 = \dfrac{5}{2\sqrt{5 x - 8}}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to multiply by the derivative of the inside → $\dfrac{1}{2\sqrt{5 x - 8}}$
• Took the derivative of $\sqrt{u}$ to be $\frac{1}{\sqrt{u}}\mathclose{,}$ losing the $\frac{1}{2}$ → $\dfrac{5}{\sqrt{5 x - 8}}$

Answer: $2 h + 21$

$\dfrac{f(4 + h) - f(4)}{h}\mathclose{.}$ Expand $f(4 + h) = 2 h^{2} + 21 h + 48$ and subtract $f(4) = 48\mathclose{:}$ the difference is $2 h^{2} + 21 h\mathclose{,}$ and every term has a factor $h\mathclose{.}$ Divided by $h\mathclose{:}$ $2 h + 21\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not divide by $h$ → $2 h^{2} + 21 h$
• Expanded $(4 + h)^2$ as $16 + h^2$ → $2 h + 5$

Answer: $x\left(2\sin(5x) + 5x\cos(5x)\right)$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{2}$ and $v = \sin(5x)\mathclose{:}$ $u' = 2x$ and $v' = 5\cos(5x)\mathclose{.}$ So $f'(x) = 2x\sin(5x) + x^{2} \cdot 5\cos(5x)\mathclose{,}$ which factorises as $x\left(2\sin(5x) + 5x\cos(5x)\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $10x\cos(5x)$
• Forgot the factor $5$ that the chain rule gives inside $\sin(5x)$ → $x\left(2\sin(5x) + x\cos(5x)\right)$

Answer: $4\cos(x) - 3\sin(x)$

$\sin(x)$ becomes $\cos(x)$ and $\cos(x)$ becomes $-\sin(x)\mathclose{;}$ the numbers in front stay. So $f'(x) = 4\cos(x) - 3\sin(x)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the derivative of $\cos(x)$ to be $+\sin(x)$ → $4\cos(x) + 3\sin(x)$
• Took the derivative of $\sin(x)$ to be $-\cos(x)$ → $-4\cos(x) - 3\sin(x)$

Answer: $\left(8 x - 4\right)e^{4 x^{2} - 4 x}$

Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside: $e^{4 x^{2} - 4 x} \cdot \left(8 x - 4\right) = \left(8 x - 4\right)e^{4 x^{2} - 4 x}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to multiply by the derivative of the exponent → $e^{4 x^{2} - 4 x}$
• Brought the exponent down like a power, $u\,e^{u - 1}$ → $\left(4 x^{2} - 4 x\right)e^{4 x^{2} - 4 x - 1}$

Answer: $2e^{x} - 4\ln(5) \cdot 5^{x}$

$e^{x}$ is its own derivative, and $5^{x}$ has the derivative $5^{x}\ln(5)\mathclose{.}$ So $f'(x) = 2e^{x} - 4\ln(5) \cdot 5^{x}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Treated $5^{x}$ like a power of $x\mathclose{,}$ as $x \cdot 5^{x - 1}$ → $2e^{x} - 4x \cdot 5^{x - 1}$
• Forgot the factor $\ln(5)$ → $2e^{x} - 4 \cdot 5^{x}$

Answer: $-3$

The difference quotient is the change in $f$ divided by the change in $x\mathclose{:}$ $\dfrac{f(3) - f(0)}{3 - 0} = \dfrac{-8 - 1}{3} = -3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not divide by the length of the interval → $-9$
• Subtracted the $x$-values the other way round → $3$

Answer: $x^{2}\left(3\cos(2x) - 2x\sin(2x)\right)$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{3}$ and $v = \cos(2x)\mathclose{:}$ $u' = 3x^{2}$ and $v' = -2\sin(2x)\mathclose{.}$ So $f'(x) = 3x^{2}\cos(2x) + x^{3} \cdot \left(-2\sin(2x)\right)\mathclose{,}$ which factorises as $x^{2}\left(3\cos(2x) - 2x\sin(2x)\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $-6x^{2}\sin(2x)$
• Forgot the factor $2$ that the chain rule gives inside $\cos(2x)$ → $x^{2}\left(3\cos(2x) - x\sin(2x)\right)$
• Took the derivative of $\cos$ to be $+\sin$ → $x^{2}\left(3\cos(2x) + 2x\sin(2x)\right)$

Answer: $50 x^{2} + 80 x + 23$

Put $g(x)$ in place of every $x$ in the outer function: $f(g(x)) = 2\left(5 x + 4\right)^{2} - 9\mathclose{,}$ and $\left(5 x + 4\right)^{2} = 25 x^{2} + 40 x + 16\mathclose{.}$ Altogether $f(g(x)) = 50 x^{2} + 80 x + 23\mathclose{.}$

Common mistakes, and the answer each one gives:

• Composed the wrong way round, working out $g(f(x))$ → $10 x^{2} - 41$
• Squared each term of $g(x)$ separately → $50 x^{2} + 23$
• Multiplied $f(x)$ by $g(x)$ instead of composing them → $10 x^{3} + 8 x^{2} - 45 x - 36$

Answer: $\dfrac{- 6 x^{2} - 4 x + 6}{\left(x^{2} + 1\right)^{2}}$

The quotient rule is $\dfrac{u'v - uv'}{v^2}$ — and the ORDER of the two products matters, unlike the product rule. With $u = 6 x + 2$ and $v = x^{2} + 1\mathclose{:}$ $u' = 6$ and $v' = 2x\mathclose{.}$ The numerator is $6\left(x^{2} + 1\right) - \left(6 x + 2\right) \cdot 2x = - 6 x^{2} - 4 x + 6\mathclose{,}$ so $f'(x) = \dfrac{- 6 x^{2} - 4 x + 6}{\left(x^{2} + 1\right)^{2}}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Wrote $uv' - u'v$ in the numerator, the two products swapped → $\dfrac{6 x^{2} + 4 x - 6}{\left(x^{2} + 1\right)^{2}}$
• Differentiated top and bottom separately → $\frac{3}{x}$

Answer: $\dfrac{-3}{x} + \dfrac{2}{x\ln(10)}$

$\ln(x)$ has the derivative $\dfrac{1}{x}\mathclose{.}$ And $\log_{10}(x) = \dfrac{\ln(x)}{\ln(10)}\mathclose{,}$ so its derivative is $\dfrac{1}{x\ln(10)}\mathclose{.}$ So $f'(x) = \dfrac{-3}{x} + \dfrac{2}{x\ln(10)}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot the $\ln(10)$ in the derivative of $\log_{10}(x)$ → $\dfrac{-1}{x}$
• Multiplied by $\ln(10)$ instead of dividing by it → $\dfrac{-3}{x} + \dfrac{2\ln(10)}{x}$

Answer: $40x\left(5 x^{2} - 6\right)^{3}$

Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside: $4\left(5 x^{2} - 6\right)^{3} \cdot 10x = 40x\left(5 x^{2} - 6\right)^{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to multiply by the derivative of the inside → $4\left(5 x^{2} - 6\right)^{3}$
• Differentiated the inside and raised THAT to the power → $10000 x^{4}$

Answer: $7$

On $[-1, -1 + h]$ the difference quotient simplifies to $7 - h\mathclose{.}$ As $h$ goes to $0$ that closes in on $7\mathclose{:}$ the slope at $x = -1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took $h = 1$ instead of letting $h$ go to $0$ → $6$
• Gave the function value $f(-1)$ instead → $-2$

Answer: $5x^{4}\left(x + 1\right)e^{5x}$

Product rule, $f' = u'v + uv'\mathclose{,}$ with $u = x^{5}$ and $v = e^{5x}\mathclose{:}$ $u' = 5x^{4}$ and $v' = 5e^{5x}\mathclose{.}$ So $f'(x) = 5x^{4}e^{5x} + x^{5} \cdot 5e^{5x}\mathclose{,}$ which factorises as $5x^{4}\left(x + 1\right)e^{5x}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the two derivatives, $u' \cdot v'$ → $25x^{4}e^{5x}$
• Forgot the factor $5$ that the chain rule gives inside $e^{5x}$ → $x^{4}\left(x + 5\right)e^{5x}$

Answer: $18 x^{5} + 6 x - 6$

Term by term (the sum rule): $x^n$ becomes $nx^{n-1}\mathclose{,}$ a number in front stays in front, and the constant $-8$ becomes $0\mathclose{.}$ So $f'(x) = 18 x^{5} + 6 x - 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Kept the constant $-8$ instead of making it $0$ → $18 x^{5} + 6 x - 14$
• Multiplied by each power but did not lower it → $18 x^{6} + 6 x^{2} - 6 x$