Differentiation rules: chain, product, quotient

Worked solutions

Problem 1

1. $u = x^3$, $v = e^x$; $u' = 3x^2$, $v' = e^x$.The function is a product of two factors whose derivatives are in the table, so the product rule applies.
2. $(x^3e^x)' = 3x^2 \cdot e^x + x^3 \cdot e^x$.Product rule, $u'v + uv'$: each term differentiates one factor and leaves the other alone.
3. $e^x(x^3 + 3x^2)$Both terms contain $e^x$, so it factors out.

Problem 2

1. $u = x$, $v = 1 + x^2$; $u' = 1$, $v' = 2x$.The function is a quotient, so name the numerator and the denominator and differentiate each.
2. $\Big(\dfrac{x}{1+x^2}\Big)' = \dfrac{1\cdot(1+x^2) - x\cdot 2x}{(1+x^2)^2}$.Quotient rule: $u'v$ first, $uv'$ subtracted, over $v^2$.
3. $1\cdot(1+x^2) - x\cdot 2x = 1 + x^2 - 2x^2 = 1 - x^2$.Expand the numerator and collect the $x^2$ terms.
4. $\dfrac{1 - x^2}{(1+x^2)^2}$Sanity check: the function rises to its maximum $\tfrac12$ at $x = 1$ and falls after it, and the derivative is $0$ at $x = 1$, positive for $|x| < 1$ and negative for $x > 1$.

Problem 3

1. Outer function $\sin$, inner function $g(x) = 3x^2$.$\sin$ is applied to $3x^2$, not to $x$, so this is a composition and needs the chain rule.
2. The derivative of $\sin$, evaluated at the inner function: $\cos(3x^2)$.$(\sin t)' = \cos t$, and the chain rule evaluates it at $t = 3x^2$.
3. The inner derivative: $g'(x) = 6x$.Power rule with the constant 3 carried along.
4. $6x\cos(3x^2)$Chain rule, $f'(g)\,g'$: the product of steps 2 and 3, with the factor $6x$ written first.

Problem 4

1. Outer function $\ln$, inner function $1 + e^x$.The log is applied to $1 + e^x$, not to $x$, so the chain rule applies.
2. The derivative of $\ln$, evaluated at the inner function: $\dfrac{1}{1+e^x}$.$(\ln t)' = 1/t$ at $t = 1 + e^x$, which is positive for every $x$.
3. The inner derivative: $(1 + e^x)' = e^x$.The constant 1 has derivative 0.
4. $\dfrac{d}{dx}\ln(1+e^x) = \dfrac{e^x}{1+e^x}$.Chain rule: step 2 times step 3.
5. $\dfrac{e^x}{1+e^x} = \dfrac{1}{e^{-x} + 1}$.Divide top and bottom by $e^x$, which is never 0, so that $x$ appears only as $e^{-x}$, the form in the definition of $\sigma$.
6. $\sigma(x)$$1/(1 + e^{-x})$ is the definition of $\sigma(x)$: softplus is an antiderivative of the sigmoid.

Problem 5

1. $\sigma = (1+e^{-x})^{-1}$.Written as a power of an inner function, $\sigma$ needs only the chain rule and the power rule, not the quotient rule.
2. The derivative of $t^{-1}$, evaluated at the inner function: $-(1+e^{-x})^{-2}$.$(t^{-1})' = -t^{-2}$, at $t = 1 + e^{-x}$.
3. The inner derivative: $(1 + e^{-x})' = e^{-x}\cdot(-1) = -e^{-x}$.$e^{-x}$ is itself a composition: the derivative of $e^t$ at $t = -x$ is $e^{-x}$, and $(-x)' = -1$.
4. $\sigma'(x) = -(1+e^{-x})^{-2}\cdot(-e^{-x}) = \dfrac{e^{-x}}{(1+e^{-x})^2}$.Chain rule: step 2 times step 3; the two minus signs cancel.
5. $\dfrac{e^{-x}}{(1+e^{-x})^2} = \sigma(x)\cdot\sigma(x)\,e^{-x}$.Each factor $1/(1+e^{-x})$ is $\sigma(x)$, so the square in the denominator gives $\sigma(x)^2$.
6. $e^{-x} = \dfrac{1}{\sigma(x)} - 1$.$1/\sigma(x) = 1 + e^{-x}$ by the definition of $\sigma$; subtract 1.
7. $\sigma(x)^2\Big(\dfrac{1}{\sigma(x)} - 1\Big) = \sigma(x) - \sigma(x)^2$.Substitute step 6 into step 5 and multiply out.
8. $\sigma'(x) = \sigma(x)\big(1-\sigma(x)\big)$Factor out $\sigma(x)$. The derivative is largest, $\tfrac14$, where $\sigma(x) = \tfrac12$, at $x = 0$.

Problem 6

1. $u = e^x - e^{-x}$, $v = e^x + e^{-x}$; $u' = e^x + e^{-x}$, $v' = e^x - e^{-x}$.$(e^{-x})' = -e^{-x}$ (Problem 5, step 3), so differentiating flips the sign of each $e^{-x}$ term: $u' = v$ and $v' = u$.
2. $\dfrac{d}{dx}\tanh x = \dfrac{(e^x+e^{-x})^2 - (e^x-e^{-x})^2}{(e^x+e^{-x})^2}$.Quotient rule: $u'v = v^2$ and $uv' = u^2$.
3. $\dfrac{v^2 - u^2}{v^2} = 1 - \Big(\dfrac{u}{v}\Big)^2$.Split the fraction into two terms; the first is 1.
4. $\dfrac{d}{dx}\tanh x = 1 - \tanh^2 x$.$u/v$ is $\tanh x$ by definition.
5. $2\sigma(2x) - 1 = \dfrac{2}{1+e^{-2x}} - 1 = \dfrac{2 - (1+e^{-2x})}{1+e^{-2x}} = \dfrac{1 - e^{-2x}}{1+e^{-2x}}$.The definition of $\sigma$ at $2x$, then put 1 over the same denominator.
6. $\dfrac{1 - e^{-2x}}{1+e^{-2x}} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}$.Multiply top and bottom by $e^x$: $e^x \cdot e^{-2x} = e^{-x}$.
7. $1 - \tanh^2 x$; and $2\sigma(2x) - 1 = \dfrac{2}{1+e^{-2x}} - 1 = \dfrac{1 - e^{-2x}}{1+e^{-2x}} = \tanh x$Steps 4 to 6. So $\tanh$ is a sigmoid stretched to the range $(-1, 1)$ and squeezed horizontally by 2.

Problem 7

1. Let $y = x^x$. Then $\ln y = x\ln x$.The exponent is the variable, so the power rule, which needs a constant exponent, does not apply; the log turns the exponent into a factor. $x > 0$ makes $y > 0$, so $\ln y$ is defined.
2. $\dfrac{d}{dx}\ln y = \dfrac{1}{y}\cdot y'$.$y$ is a function of $x$, so this is a chain rule: the derivative of $\ln$ evaluated at $y$ is $1/y$, times the inner derivative $y'$.
3. $\dfrac{d}{dx}(x\ln x) = 1\cdot\ln x + x\cdot\dfrac{1}{x} = \ln x + 1$.Product rule with $u = x$ and $v = \ln x$; $(\ln x)' = 1/x$ for $x > 0$.
4. $\dfrac{y'}{y} = \ln x + 1$, so $y' = y(\ln x + 1)$.The two sides of step 1 are equal for every $x > 0$, so their derivatives are equal; multiply by $y$.
5. $x^x(\ln x + 1)$Substitute $y = x^x$.

Problem 8

1. $\ln\dfrac{e^x}{1+e^x} = \ln e^x - \ln(1+e^x) = x - \ln(1+e^x)$.Simplify before differentiating: the log of a quotient is a difference of logs, and $\ln e^x = x$. That replaces a chain rule wrapped around a quotient rule with two terms whose derivatives are already known.
2. $\dfrac{d}{dx}\big(x - \ln(1+e^x)\big) = 1 - \sigma(x)$.$x$ has derivative 1, and the derivative of $\ln(1+e^x)$ is $\sigma(x)$ (Problem 4).
3. $1 - \sigma(x) = 1 - \dfrac{1}{1+e^{-x}} = \dfrac{e^{-x}}{1+e^{-x}}$.Put 1 over the denominator $1 + e^{-x}$.
4. $\dfrac{e^{-x}}{1+e^{-x}} = \dfrac{1}{e^x + 1}$.Divide top and bottom by $e^{-x}$.
5. $\sigma(-x)$$1/(1 + e^{x})$ is the definition of $\sigma$ evaluated at $-x$. Since $e^x/(1+e^x) = \sigma(x)$ (Problem 4, step 5), this also says $\tfrac{d}{dx}\ln\sigma(x) = 1 - \sigma(x) = \sigma(-x)$.

Problem 9

1. Four layers, outermost first: $\sqrt{t}$, then $1 + s^2$, then $\sin w$, then $w = 2x$.Reading the function from the outside in lists the compositions; the chain rule needs one factor per layer.
2. The derivative of $\sqrt{t}$, evaluated at the inner function $1 + \sin^2 2x$: $\dfrac{1}{2\sqrt{1+\sin^2 2x}}$.$(\sqrt t)' = 1/(2\sqrt t)$ at $t = 1 + \sin^2 2x$, which is at least 1, so the square root is never 0.
3. The derivative of $1 + s^2$, evaluated at $\sin 2x$: $2\sin 2x$.$(1 + s^2)' = 2s$ at $s = \sin 2x$.
4. The derivative of $\sin w$, evaluated at $2x$: $\cos 2x$; and $(2x)' = 2$.$(\sin w)' = \cos w$ at $w = 2x$, and the innermost derivative is the constant 2.
5. $\dfrac{d}{dx}\sqrt{1+\sin^2 2x} = \dfrac{2\sin 2x\cos 2x\cdot 2}{2\sqrt{1+\sin^2 2x}}$.Chain rule through all four layers: the product of the factors in steps 2 to 4.
6. $2\sin 2x\cos 2x = \sin 4x$.Double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ with $\theta = 2x$.
7. $\dfrac{\sin 4x}{\sqrt{1+\sin^2(2x)}}$Substitute step 6 into the numerator of step 5; the remaining factor 2 cancels the 2 in the denominator.

Problem 10

1. Treat $y$ as a function of $x$ and differentiate both sides of $x^2 + xy + y^2 = 7$ with respect to $x$.The equation holds at every $x$ in a stretch of the curve, so the derivatives of the two sides are equal there. Write $y'$ for $dy/dx$.
2. $\dfrac{d}{dx}x^2 = 2x$.Power rule.
3. $\dfrac{d}{dx}(xy) = 1\cdot y + x\,y'$.Product rule with $u = x$ and $v = y$; $y$ depends on $x$, so its derivative $y'$ stays.
4. $\dfrac{d}{dx}y^2 = 2y\cdot y'$.Chain rule: the derivative of $t^2$ evaluated at $y$ is $2y$, times the inner derivative $y'$.
5. $2x + y + xy' + 2yy' = 0$.Add steps 2 to 4; the right side, 7, is constant.
6. $y'(x + 2y) = -(2x + y)$.Collect the terms containing $y'$ on one side.
7. $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$, valid where $x + 2y \neq 0$Divide by $x + 2y$, which is allowed only where it is not 0. Where $x + 2y = 0$ the curve meets $y = -x/2$, so $\tfrac34x^2 = 7$ and $x = \pm\sqrt{28/3}$; the tangent there is vertical, and near those two points $y$ is not a function of $x$.