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Derivative of ln(x)
Derivative of ln(x)

Derivative of $l n (𝑥)$

Let's work some practice problems taking the derivative of the natural log, $l n (𝑥),$ each of course with a complete solution a click away.

Note that problems involving this particular derivative essentially always require the Chain Rule, so this is good additional Chain Rule practice too.

Matheno Essentials: Derivative of $l n (𝑥)$

𝑑 𝑑 𝑥 l n 𝑥 = 1 𝑥

Applying the Chain rule, we have

𝑑 𝑑 𝑥 l n (s t u f f) = 1 (s t u f f) \cdot 𝑑 𝑑 𝑥 (s t u f f)

You'll usually see this written as

𝑑 𝑑 𝑥 l n 𝑢 = 1 𝑢 \cdot 𝑑 𝑢 𝑑 𝑥

Practice Problem #1

Differentiate

𝑓 (𝑥) = l n (𝑥 2 + 3 𝑥 - 1) .

𝑓' (𝑥) = 𝑑 𝑑 𝑥 [l n (𝑥 2 + 3 𝑥 - 1)] = 1 (𝑥 2 + 3 𝑥 - 1) \cdot [𝑑 𝑑 𝑥 (𝑥 2 + 3 𝑥 - 1)] = 2 𝑥 + 3 (𝑥 2 + 3 𝑥 - 1) ✓

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Practice Problem #2

Differentiate

𝑓 (𝑥) = \sqrt l n 𝑥 2 .

𝑓' (𝑥) = 𝑑 𝑑 𝑥 [\sqrt l n 𝑥 2] = 𝑑 𝑑 𝑥 [l n 𝑥 2] 1 / 2 = 12 [l n 𝑥 2] - 1 / 2 \cdot [𝑑 𝑑 𝑥 l n 𝑥 2] = 12 1 \sqrt l n 𝑥 2 \cdot [1 𝑥 2 (𝑑 𝑑 𝑥 𝑥 2)] = 12 1 \sqrt l n 𝑥 2 \cdot 1 𝑥 2 (2 𝑥) = 1 𝑥 \sqrt l n 𝑥 2 ✓

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Practice Problem #3

Differentiate

𝑓 (𝑥) = l n (l n 𝑥) .

𝑓' (𝑥) = 𝑑 𝑑 𝑥 [l n (l n 𝑥)] = 1 l n 𝑥 \cdot 𝑑 𝑑 𝑥 (l n 𝑥) = 1 l n 𝑥 \cdot 1 𝑥 = 1 𝑥 l n 𝑥 ✓

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Practice Problem #4

Differentiate

𝑓 (𝑥) = l n (𝑥 2 𝑒 𝑥) .

We'll solve this two ways:
Method 1: Chain rule and Product rule

𝑑 𝑑 𝑥 l n (𝑥 2 𝑒 𝑥) = 1 𝑥 2 𝑒 𝑥 \cdot 𝑑 𝑑 𝑥 [𝑥 2 𝑒 𝑥] = 1 𝑥 2 𝑒 𝑥 \cdot [(𝑑 𝑑 𝑥 𝑥 2) 𝑒 𝑥 + 𝑥 2 (𝑑 𝑑 𝑥 𝑒 𝑥)] = 1 𝑥 2 𝑒 𝑥 \cdot [2 𝑥 𝑒 𝑥 + 𝑥 2 𝑒 𝑥] = 2 𝑥 𝑒 𝑥 𝑥 2 𝑒 𝑥 + 𝑥 2 𝑒 𝑥 𝑥 2 𝑒 𝑥 = 2 𝑥 + 1 ✓

Method 2: Use log property.
Recall that

l n (𝑓 * 𝑔) = l n (𝑓) + l n (𝑔) H e n c e l n (𝑓 * 𝑔) l n (𝑥 2 𝑒 𝑥) = l n (𝑥 2) + l n (𝑒 𝑥)

Then we can compute the derivative:

𝑑 𝑑 𝑥 l n (𝑥 2 𝑒 𝑥) = 𝑑 𝑑 𝑥 l n (𝑥 2) + 𝑑 𝑑 𝑥 l n (𝑒 𝑥) = 1 𝑥 2 \cdot (𝑑 𝑑 𝑥 𝑥 2) + 1 𝑒 𝑥 \cdot (𝑑 𝑑 𝑥 𝑒 𝑥) = 1 𝑥 2 \cdot 2 𝑥 + 1 𝑒 𝑥 \cdot 𝑒 𝑥 = 2 𝑥 + 1 ✓

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Practice Problem #5

Differentiate

𝑓 (𝑥) = l n (c o s 𝑥 5 𝑥) .

We'll solve this two ways.
Method 1: Chain rule and Quotient rule.

𝑑 𝑑 𝑥 [l n (c o s 𝑥 5 𝑥)] = 1 c o s 𝑥 5 𝑥 \cdot 𝑑 𝑑 𝑥 [c o s 𝑥 5 𝑥] = 5 𝑥 c o s 𝑥 \cdot ⎡ ⎢ ⎢ ⎢ ⎣ (𝑑 𝑑 𝑥 c o s 𝑥) 5 𝑥 - c o s 𝑥 (𝑑 𝑑 𝑥 5 𝑥) (5 𝑥) 2 ⎤ ⎥ ⎥ ⎥ ⎦ = 5 𝑥 c o s 𝑥 \cdot [(- s i n 𝑥) 5 𝑥 - c o s 𝑥 (5) (5 𝑥) 2] = 1 c o s 𝑥 [- 5 𝑥 s i n 𝑥 - 5 c o s 𝑥 5 𝑥] = - 5 𝑥 s i n 𝑥 5 𝑥 c o s 𝑥 - 5 c o s 𝑥 5 𝑥 c o s 𝑥 = - t a n 𝑥 - 1 𝑥 ✓

Method 2: Use log property
Recall that

l n (𝑓 𝑔) = l n (𝑓) - l n (𝑔) H e n c e l n (𝑓 * 𝑔) l n (c o s 𝑥 5 𝑥) = l n (c o s 𝑥) - l n (5 𝑥)

Then we can compute the derivative:

𝑑 𝑑 𝑥 [l n (c o s 𝑥 5 𝑥)] = 𝑑 𝑑 𝑥 l n (c o s 𝑥) - 𝑑 𝑑 𝑥 l n (5 𝑥) = 1 c o s 𝑥 \cdot (𝑑 𝑑 𝑥 c o s 𝑥) - 1 5 𝑥 \cdot (𝑑 𝑑 𝑥 5 𝑥) = 1 c o s 𝑥 \cdot (- s i n 𝑥) - 1 5 𝑥 \cdot (5) = - t a n 𝑥 - 1 𝑥 ✓

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E.2 Derivative of Inverse Functions

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Derivative Rules: Practice Problems & Solutions

Derivative of ln⁡(𝑥)

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Derivative of $l n (𝑥)$