Porisma

Worked solution

Take logarithms of a positive variable power

Given that y=(2x+1)xy=(2x+1)^x for x>0x>0, find dydx\dfrac{dy}{dx}. marks

[2012/Promo/CJC/11(b)]

CJC · Review Problems · Question 10

Answer

(2x+1)x[2x/(2x+1)+ln⁡(2x+1)](2x+1)^x[2x/(2x+1)+\ln(2x+1)].

Logarithmic differentiation

Take logarithms: ln⁡y=xln⁡(2x+1)\ln y=x\ln(2x+1). The product rule gives y′/y=2x/(2x+1)+ln⁡(2x+1)y\prime/y=2x/(2x+1)+\ln(2x+1). Substitute back to obtain dydx=(2x+1)x[2x2x+1+ln⁡(2x+1)]\dfrac{dy}{dx}=(2x+1)^x\left[\dfrac{2x}{2x+1}+\ln(2x+1)\right].

How did this part go?

Common mistakes

Treat the logarithm of the power as a product; neither the base nor exponent is constant.

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