Risolvere: $\frac{d}{dx}\left(\frac{x\sqrt{x^2+9}}{\sqrt[3]{\left(x+4\right)^{2}}}\right)$
Esercizio
$\frac{dy}{dx}\left(\frac{x\sqrt{x^2+9}}{\left(x+4\right)^{\frac{2}{3}}}\right)$
Soluzione passo-passo
Impara online a risolvere i problemi di limiti per sostituzione diretta passo dopo passo. Find the derivative d/dx((x(x^2+9)^(1/2))/((x+4)^(2/3))). 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{x\sqrt{x^2+9}}{\sqrt[3]{\left(x+4\right)^{2}}}\right) e x=\frac{x\sqrt{x^2+9}}{\sqrt[3]{\left(x+4\right)^{2}}}. Applicare la formula: y=x\to \ln\left(y\right)=\ln\left(x\right), dove x=\frac{x\sqrt{x^2+9}}{\sqrt[3]{\left(x+4\right)^{2}}}. Applicare la formula: y=x\to y=x, dove x=\ln\left(\frac{x\sqrt{x^2+9}}{\sqrt[3]{\left(x+4\right)^{2}}}\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=\ln\left(x\right)+\frac{1}{2}\ln\left(x^2+9\right)- \left(\frac{2}{3}\right)\ln\left(x+4\right).
Find the derivative d/dx((x(x^2+9)^(1/2))/((x+4)^(2/3)))
Risposta finale al problema
$\left(\frac{1}{x}+\frac{x}{x^2+9}+\frac{-2}{3\left(x+4\right)}\right)\frac{x\sqrt{x^2+9}}{\sqrt[3]{\left(x+4\right)^{2}}}$