Risolvere: $\frac{d}{dx}\left(\frac{\left(2x^2+3\right)^2}{\left(x+1\right)^2\left(5x+6\right)}\right)$
Esercizio
$\frac{dy}{dx}\left(\frac{\left(2x^2+3\right)^2}{\left(x+1\right)^2\left(5x+6\right)}\right)$
Soluzione passo-passo
Impara online a risolvere i problemi di identità trigonometriche passo dopo passo. Find the derivative d/dx(((2x^2+3)^2)/((x+1)^2(5x+6))). Applicare la formula: \frac{d}{dx}\left(x\right)=y=x, dove d/dx=\frac{d}{dx}, d/dx?x=\frac{d}{dx}\left(\frac{\left(2x^2+3\right)^2}{\left(x+1\right)^2\left(5x+6\right)}\right) e x=\frac{\left(2x^2+3\right)^2}{\left(x+1\right)^2\left(5x+6\right)}. Applicare la formula: y=x\to \ln\left(y\right)=\ln\left(x\right), dove x=\frac{\left(2x^2+3\right)^2}{\left(x+1\right)^2\left(5x+6\right)}. Applicare la formula: y=x\to y=x, dove x=\ln\left(\frac{\left(2x^2+3\right)^2}{\left(x+1\right)^2\left(5x+6\right)}\right) e y=\ln\left(y\right). Applicare la formula: \ln\left(y\right)=x\to \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\right), dove x=2\ln\left(2x^2+3\right)-2\ln\left(x+1\right)-\ln\left(5x+6\right).
Find the derivative d/dx(((2x^2+3)^2)/((x+1)^2(5x+6)))
Risposta finale al problema
$\left(\frac{8x}{2x^2+3}+\frac{-2}{x+1}+\frac{-5}{5x+6}\right)\frac{\left(2x^2+3\right)^2}{\left(x+1\right)^2\left(5x+6\right)}$