อนุพันธ์

$ y= (x^3+x^2+1)^4 (4x+5)^6 \sin (x^2+3) $

วิธีทำ

\begin{align} & \cssId{Step1}{ take \quad \ln \quad จะได้ \quad \ln y= 4 \ln (x^3+x^2+1) + 6 \ln (4x+5) +\ln ( \sin (x^2+3) ) } \\ &\cssId{Step2}{ดังนั้น \quad \frac{1}{y} \frac{dy}{dx} = 4 \frac{1}{x^3+x^2+1} \frac{d}{dx} (x^3+x^2+1) +6 \frac{1}{4x+5} \frac{d}{dx} (4x+5) +\frac{1}{\sin (x^2+3)} \frac{d}{dx} \sin (x^2+3) }\\ &\cssId{Step3}{ = 4 \frac{1}{x^3+x^2+1} (3x^2+2x) +6 \frac{1}{4x+5} (4) +\frac{1}{\sin (x^2+3)} \cos (x^2+3) (2x) } \\ &\cssId{Step4}{ ดังนั้น \quad \frac{dy}{dx} = y ( \frac{4(3x^2+2x) }{x^3+x^2+1} + \frac{24}{4x+5} +\frac{2x \cos (x^2+3) }{\sin (x^2+3)} ) } \\ &\cssId{Step5}{ = (x^3+x^2+1)^4 (4x+5)^6 \sin (x^2+3) ( \frac{4(3x^2+2x) }{x^3+x^2+1} + \frac{24}{4x+5} +\frac{2x \cos (x^2+3) }{\sin (x^2+3)} ) } \\ \end{align}