A differentiable function is defined for all with

MATHEMATICS · Single correct · AP · Calculus AB · Differentiation: Composite, Implicit, and Inverse Functions · Differentiation: Composite, Implicit, and Inverse Functions

Problem

A differentiable function f(x)=(g(x)h(x))k(x)f(x)=\bigl(g(x)h(x)\bigr)^{k(x)} is defined for all xx with g(x),h(x),k(x)>0g(x),h(x),k(x)>0. Which of the following statements best explains why setting G(x)=ln⁡f(x)G(x)=\ln f(x) can make it easier to find f′(x)f'(x)? (a) Because taking ln⁡\ln turns the complicated form into a sum of three separate functions, whose derivative is simply the sum g′(x)+h′(x)+k′(x)g'(x)+h'(x)+k'(x). (b) The natural logarithm transforms products into sums and variable powers into products, so differentiating GG requires only sums of simpler terms. (c) Because ln⁡\ln "cancels" the exponent so that G(x)=k(x)G(x)=k(x), which means f′(x)=k′(x)f'(x)=k'(x). (d) Since ddxln⁡x=1\dfrac{d}{dx}\ln x = 1, differentiating GG produces the original inside unchanged, giving f′(x)=g(x)h(x)f'(x)=g(x)h(x).

Related theory

Browse the problem library