Assertion (A): d/dx 3^x = x · 3^x - 1. Reason (R): d/dx a^x = a^xlog a for a > 0.

Question 57

EasyOriginal question

Assertion (A): $\frac{d}{dx} 3^{x} = x \cdot 3^{x - 1}$ .
Reason (R): $\frac{d}{dx} a^{x} = a^{x}\log a$ for $a > 0$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

R is the standard result: $a^{x} = e^{x\log a}$ , whose derivative is $e^{x\log a}\log a = a^{x}\log a$ .
A is false: here the variable is in the power, so $\frac{d}{dx} 3^{x} = 3^{x}\log 3$ , not the power rule.

Worth noting

A applies the power rule $nx^{n - 1}$ , which needs a variable base and a constant power.

Concept: An exponential or a logarithm to a base other than e
Typical timeTypical: 60 s
Your timeYours: —

The rule this testsExponential and logarithmic functions

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Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.

Assertion (A): ddx3x=x⋅3x−1\frac{d}{dx} 3^{x} = x \cdot 3^{x - 1}dxd​3x=x⋅3x−1.Reason (R): ddxax=axlog⁡a\frac{d}{dx} a^{x} = a^{x}\log adxd​ax=axloga for a>0a > 0a>0.

Explanation

Assertion (A): $\frac{d}{dx} 3^{x} = x \cdot 3^{x - 1}$ .
Reason (R): $\frac{d}{dx} a^{x} = a^{x}\log a$ for $a > 0$ .