Practice problem
Find the derivative. f(x)=ln(x2+3x−5)f(x)=\ln(x^2+3x-5)f(x)=ln(x2+3x−5)
f′(x)=1x2+3x−5f'(x)=\frac{1}{x^2+3x-5}f′(x)=x2+3x−51
f′(x)=2x+3x2+3x−5f'(x)=\frac{2x+3}{x^2+3x-5}f′(x)=x2+3x−52x+3
f′(x)=12x+3f'(x)=\frac{1}{2x+3}f′(x)=2x+31
f′(x)=x2+3x−52x+3f'(x)=\frac{x^2+3x-5}{2x+3}f′(x)=2x+3x2+3x−5
Type a concept, like “slope” or “photosynthesis”.
