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Basic Exercises on Differentiation Rules and Structural Recognitionmd 95ac6d4
exercise/math/calculus/differentiation-rules-and-structure-recognition.exercise.n.md

Basic Exercises on Differentiation Rules and Structural Recognition

date2026-07-15document_iddoc_4045cb58687d413ecd1201a33b27f349description積・商・連鎖律・対数微分を、関数の構造判定と適用条件から確認する基本演習である。prerequisites微分公式と計算法 / 導関数の定義と差商type問題演習content_typeexercisestatusactiverelateddata/lecture/math/calculus/differentiation-rules-and-computation.lecture.n.md / data/exercise/math/calculus/differentiation-and-derivative-computation.exercise.n.md
mathcalculusexercisedifferentiation-rules
data/lecture/math/calculus/differentiation-rules-and-computation.lecture.n.md

1Problem 1

Differentiate (x2+1)sinx.

1.1Sample Solution

By the product rule, the derivative is

2xsinx+(x2+1)cosx.

1.2Explanation

Both factors contribute to the change of a product: (fg)=fg+fg, not fg.


2Problem 2

Differentiate [PARSE ERROR: Undefined("Command(\"dfrac\")")]x2+1x-1.

2.1Sample Solution

The domain is x1. By the quotient rule,

2x(x-1)-(x2+1)(x-1)2=x2-2x-1(x-1)2.

2.2Explanation

The domain of the original function is R{1}. This restriction remains in force after deriving a formula for the derivative.


3Problem 3

Differentiate ex2cosx.

3.1Sample Solution

The product rule and chain rule give

2xex2cosx-ex2sinx.

3.2Explanation

The derivative of the inner function x2 must multiply the derivative of the outer exponential function.


4Problem 4

For x>0, differentiate f(x)=xx.

4.1Sample Solution

Set y=xx. Since x>0, taking logarithms gives

lny=xlnx.

Differentiating both sides with respect to x yields

yy=lnx+1.

Therefore,

f(x)=xx(lnx+1)(x>0).

4.2Explanation

Both the base and exponent are variable, so neither the ordinary power rule nor the exponential rule applies directly. Logarithmic differentiation converts the exponent into a product before differentiation.

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