Math · Calculus I · Worked example

Differentiate a power and an exponential

Differentiate $f (x) = x^{4} + 4^{x}$ , then find the slope of its graph at $x = 1$ .

f (x) = x^{4} + 4^{x}

Tell the two terms apart

Both terms hold a $4$ and an $x$ , in opposite places. In $x^{4}$ the base is the variable and the exponent is fixed, so it is a power. In $4^{x}$ the exponent is the variable, so it is an exponential with base $4$ .

Differentiate the power

The power rule brings the exponent $4$ down and lowers it by one.

\frac{d}{d x} x^{4} = 4 x^{3}

Differentiate the exponential

An exponential with base $b$ has derivative $b^{x} ln b$ . Here $b = 4$ , so the derivative of $4^{x}$ is $4^{x} ln 4$ . The power rule would give $x \cdot 4^{x - 1}$ , which is wrong.

\frac{d}{d x} 4^{x} = 4^{x} ln 4

Add the two derivatives

The derivative of a sum is the sum of the derivatives, and both formulas hold for every $x$ .

f^{'} (x) = 4 x^{3} + 4^{x} ln 4

Find the slope at $x = 1$

At $x = 1$ the first term is $4$ and the second is $4 ln 4 \approx 5.545$ . The slope is $4 + 4 ln 4 \approx 9.545$ .

f^{'} (1) = 4 + 4 ln 4 \approx 9.545

Compare the two terms

At $x = 1$ the exponential already has the steeper slope, and the gap only widens. At $x = 10$ the power’s slope is $4 \times 10^{3} = 4000$ , while the exponential’s is $4^{10} ln 4$ , about $1.45$ million.

Result

$f^{'} (x) = 4 x^{3} + 4^{x} ln 4$ , and $f^{'} (1)$ $= 4 + 4 ln 4$ $\approx 9.545$ .

f^{'} (x) = 4 x^{3} + 4^{x} ln 4

Your turn

1. One like the example

Differentiate $f (x) = x^{3} - 3^{x}$ and find $f^{'} (0)$ .

Show the answer and explanation

$f^{'} (x) = 3 x^{2} - 3^{x} ln 3$ , and $f^{'} (0) = - ln 3 \approx - 1.099$ .

The power $x^{3}$ has derivative $3 x^{2}$ . The exponential $3^{x}$ has derivative $3^{x} ln 3$ . At $x = 0$ the power’s slope is $0$ and $3^{0} = 1$ , which leaves $f^{'} (0) = - ln 3$ .

\begin{array}{l} f (x) = x^{3} - 3^{x} \\ f^{'} (x) = 3 x^{2} - 3^{x} ln 3 \\ f^{'} (0) = - ln 3 \end{array}

2. A variation

Differentiate $f (x) = e^{x} + x^{e}$ , and find $f^{'} (1)$ .

Show the answer and explanation

$f^{'} (x) = e^{x} + e \cdot x^{e - 1}$ for $x > 0$ , and $f^{'} (1) = 2 e \approx 5.437$ .

The exponential $e^{x}$ is its own derivative. The power $x^{e}$ has the constant exponent $e \approx 2.718$ , so the power rule gives $e \cdot x^{e - 1}$ . At $x = 1$ both terms equal $e$ . Chalk checks the derivative row and the value $2 e$ .

\begin{array}{l} f (x) = e^{x} + x^{e} \\ f^{'} (x) = e^{x} + e x^{e - 1} \\ f^{'} (1) = 2 e \end{array}

3. Somewhere new

A culture starts with $500$ bacteria and doubles every hour, so $N (t) = 500 \cdot 2^{t}$ after $t$ hours. How fast is it growing at $t = 3$ ?

Show the answer and explanation

$N^{'} (3) = 4000 ln 2 \approx 2773$ bacteria per hour.

$N$ is a constant times an exponential with base $2$ , so $N^{'} (t) = 500 \cdot 2^{t} ln 2$ . At $t = 3$ the population is $4000$ , and the rate is $4000 ln 2$ $\approx 2773 per hour$ . At every moment the rate is $ln 2 \approx 0.693$ times the current population.

\begin{array}{l} N (t) = 500 \cdot 2^{t} \\ N^{'} (t) = 500 \cdot 2^{t} ln 2 \\ N^{'} (3) = 4000 ln 2 \end{array}

Keep exploring

Open the steps in Math: the Checker marks every row ✓. Write $4 x^{3} + x \cdot 4^{x - 1}$ as the derivative instead and it reads ✗, because $4^{x}$ is not a power of $x$ .

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Tell the two terms apart

Differentiate the power

Differentiate the exponential

Add the two derivatives

Find the slope at x=1x=1

Compare the two terms

Result

Your turn

1. One like the example

f′(x)=3x2−3xln3f′(x)=3x2−3xln3, and f′(0)=−ln3≈−1.099f′(0)=−ln3≈−1.099.

2. A variation

f′(x)=ex+e⋅xe−1f′(x)=ex+e⋅xe−1 for x>0x>0, and f′(1)=2⁢e≈5.437f′(1)=2e≈5.437.

3. Somewhere new

N′(3)=4000ln2≈2773N′(3)=4000ln2≈2773 bacteria per hour.

Keep exploring

Study it, then try it

Find the slope at $x = 1$

$f^{'} (x) = 3 x^{2} - 3^{x} ln 3$ , and $f^{'} (0) = - ln 3 \approx - 1.099$ .

$f^{'} (x) = e^{x} + e \cdot x^{e - 1}$ for $x > 0$ , and $f^{'} (1) = 2 e \approx 5.437$ .

$N^{'} (3) = 4000 ln 2 \approx 2773$ bacteria per hour.