Product and Quotient Rules I

By the definition of the linear approximation, $L$ is the tangent line to $f$ when $x = 4$. Substitute $a = 4$ into the formula for the linear approximation:

Rewrite $\sqrt{x}$ as $x^{1/2}$. The power rule gives $f'(x) = \tfrac{1}{2}x^{-1/2}$. Evaluate $f$ and $f'$ at 4:

Substitute both values into $L(x)$:

Estimate $\sqrt{4.1}$ with $L$ at $x = 4.1$:

For comparison, R computes the true value with sqrt(4.1): $\sqrt{4.1} \approx 2.024846$. The error $f(4.1) - L(4.1) \approx 2.024846 - 2.025 = -0.000154$ is small, because 4.1 is close to $a = 4$.

Estimate $\sqrt{16}$ with $L$ at $x = 16$:

The true value is $\sqrt{16} = 4$. The error $f(16) - L(16) = 4 - 5 = -1$ is far larger in size, because 16 is far from $a = 4$. On the graph, the red line $L$ touches the curve at $(4, 2)$ and lies close to the curve for inputs near $x = 4$. At $x = 16$ the line lies above the curve.

Build the linear approximation $L(x) = f(1) + f'(1)(x - 1)$ as in Example 1, with $f(1)$ and $f'(1)$ from R. Then compare $L(1.1)$ with $f(1.1)$, and plot $f$ with $L$. Used R.

f = makeFun(x^3 - 5*x ~ x)
df = D(f(x) ~ x)
f(1)
[1] -4
df(1)
[1] -2
L = makeFun(f(1) + df(1)*(x - 1) ~ x)
L(1.1)
[1] -4.2
f(1.1)
[1] -4.169
plotFun(f(x) ~ x, xlim = range(-3, 3))
plotFun(L(x) ~ x, add = TRUE, col = "red")

The code stores the derivative $f'$ under the name df. R prints $f(1) = -4$ and $f'(1) = -2$. The linear approximation is therefore $L(x) = -4 - 2(x - 1)$. The command that defines L is the formula $L(x) = f(1) + f'(1)(x - 1)$ written in R. The error of the estimate is $f(1.1) - L(1.1) = -4.169 - (-4.2) = 0.031$. In the plot, the red line $L$ touches the black curve $f$ at $(1, f(1)) = (1, -4)$ with slope $f'(1) = -2$.

Compute $g(2)$ from g and $g'(2)$ from the derivative that D() returns, build $L$ as in Example 2, and evaluate $L(1.9)$. Used R.

g = makeFun(x^2*sin(x)/(x + 1) ~ x)
dg = D(g(x) ~ x)
g(2)
[1] 1.212397
dg(2)
[1] 0.2534019
L = makeFun(g(2) + dg(2)*(x - 2) ~ x)
L(1.9)
[1] 1.187056

To three decimals, $g(2) \approx 1.212$ and $g'(2) \approx 0.253$. The linear approximation is

and the estimate is $g(1.9) \approx L(1.9) \approx 1.187$.

By the product rule, the derivative of a product of two differentiable factors $u(x)$ and $v(x)$ is

Here the factors are $u(x) = 3x^2$, with derivative $6x$, and $v(x) = e^x$, with derivative $e^x$. Substitute them into the product rule:

The two factors are $\sqrt{t} = t^{1/2}$, with derivative $\tfrac{1}{2}t^{-1/2}$ by the power rule, and $\sin(t)$, with derivative $\cos(t)$. By the product rule,

for every $t$ with $t > 0$, where both factors are differentiable.

By the quotient rule, the derivative of a quotient of a differentiable $u(x)$ by a differentiable $v(x)$, at each input $x$ with $v(x) \neq 0$, is

Here the numerator is $u(x) = \sin(x)$, with derivative $\cos(x)$, and the denominator is $v(x) = x^2$, with derivative $2x$. Substitute them into the quotient rule, then divide the numerator and the denominator by $x$:

for every $x$ with $x \neq 0$.

The numerator is $x^2$, with derivative $2x$, and the denominator is $x + 1$, with derivative 1. By the quotient rule,

for every $x$ with $x \neq -1$.

The numerator is $\cos(x)$, with derivative $-\sin(x)$, and the denominator is $\sin(x)$, with derivative $\cos(x)$. By the quotient rule,

for every $x$ with $\sin(x) \neq 0$, because $\sin^2(x) + \cos^2(x) = 1$ for every input $x$.

The numerator $x^2\cos(x)$ is a product. By the product rule, its derivative is

The denominator is $\ln(x)$, with derivative $\frac{1}{x}$. By the quotient rule,

for every $x$ with $x > 0$ and $x \neq 1$.

• Differentiate $x^3e^x$ and $(x^2 + 1)\ln(x)$ with the product rule, because each is a product of two non-constant factors, and no algebra rewrites it as a linear combination of power functions and basic modeling functions (Lesson 29).
• Differentiate $\frac{e^x}{x}$ with the quotient rule, because its denominator is not constant, and no algebra rewrites the quotient as a linear combination of power functions and basic modeling functions.
• Differentiate $4\sin(x)$ with only the basic derivative rules, because the function is the constant 4 times $\sin(x)$. Its derivative is $4\cos(x)$.
• Differentiate $\frac{x^2 + 1}{3}$ with only the basic derivative rules, because the function is the constant $\frac{1}{3}$ times $x^2 + 1$. Its derivative is $\frac{2x}{3}$.
• Differentiate $\frac{x^5}{x^2}$ with only the basic derivative rules, because the function simplifies to $x^3$. Its derivative is $3x^2$. The simplified form $x^3$ equals $\frac{x^5}{x^2}$ only for $x \neq 0$, the domain of $\frac{x^5}{x^2}$. The derivative $3x^2$ applies on that domain.

Define $f$, differentiate it with D(), and print the derivative. Used R.

f = makeFun(3*x^2*exp(x) ~ x)
df = D(f(x) ~ x)
df
function (x)
3 * (x^2 * exp(x) + 2 * x * exp(x))

R prints the derivative as a function of $x$, with its formula on the second line. The printed form $3(x^2e^x + 2xe^x) = 3x^2e^x + 6xe^x$ matches the hand derivative. As a spot check at $x = 1$, the hand derivative is

Evaluate the R derivative and $9e$. Used R.

df(1)
[1] 24.46454
9*exp(1)
[1] 24.46454

The two outputs are equal at $x = 1$, consistent with the algebraic match.

Define $r$ with parentheses around the denominator, differentiate it with D(), and print the derivative. Used R.

r = makeFun(x^2/(x + 1) ~ x)
dr = D(r(x) ~ x)
dr
function (x)
{
.e1 <- 1 + x
x * (2 - x/.e1)/.e1
}

The line .e1 <- 1 + x stores the denominator $x + 1$ under the name .e1, and the last line is the derivative written with that name:

The printed form matches the hand derivative. As a spot check at $x = 1$, the hand derivative is

Evaluate the R derivative at 1. Used R.

dr(1)
[1] 0.75

The R value and the hand value are equal at $x = 1$, consistent with the algebraic match.