Porisma

Worked solution

Differentiate a product with constant-e exponent and a variable power

Differentiate the following with respect to xx, leaving your answers in terms of xx.

(a) eexcot⁡(x2)e^{ex}\cot\left(\dfrac x2\right), marks

(b) (2x)1/x(2x)^{1/x}. marks

[2010/ACJC/Promo/5]

CJC · Review Problems · Question 6

Answer

(a) eex[ecot⁡(x/2)−12csc⁡2(x/2)]e^{ex}[e\cot(x/2)-\frac12\csc^2(x/2)]; (b) (2x)1/x[1−ln⁡(2x)]/x2(2x)^{1/x}[1-\ln(2x)]/x^2.

(a)

The exponent is the product exex, with ee constant. The product rule gives eex[−12csc⁡2(x/2)]+cot⁡(x/2)eexee^{ex}[-\frac12\csc^2(x/2)]+\cot(x/2)e^{ex}e. Factor to obtain eex[ecot⁡(x/2)−12csc⁡2(x/2)]e^{ex}\left[e\cot(x/2)-\frac12\csc^2(x/2)\right]. The cotangent requires x≠2kπx\ne2k\pi, k∈Zk\in\mathbb Z.

How did this part go?

(b)

For x>0x>0, let y=(2x)1/xy=(2x)^{1/x} and take logs: ln⁡y=x−1ln⁡(2x)\ln y=x^{-1}\ln(2x). Thus y′/y=x−1(1/x)−x−2ln⁡(2x)=[1−ln⁡(2x)]/x2y\prime/y=x^{-1}(1/x)-x^{-2}\ln(2x)=[1-\ln(2x)]/x^2. Therefore y′=(2x)1/xx2[1−ln⁡(2x)]y\prime=\dfrac{(2x)^{1/x}}{x^2}[1-\ln(2x)].

How did this part go?

Common mistakes

The exponent in (a) is ex, not e raised to x; in (b) both the base and exponent vary.

Solution notes

A higher-resolution source rendering confirms the exponent in (a) is the product ex. The source factor e in its derivative is correct.

Read the hint