Risolvere: $\frac{d}{dx}\left(y=\mathrm{tanh}\left(x\right)^2\ln\left(\mathrm{sech}\left(y\right)\right)\right)$
Esercizio
$\frac{dy}{dx}\left(y=ln\left(sech\right)tanh^2x\right)$
Soluzione passo-passo
Impara online a risolvere i problemi di passo dopo passo. d/dx(y=ln(sech(y))tanh(x)^2). Applicare la formula: \frac{d}{dx}\left(a=b\right)=\frac{d}{dx}\left(a\right)=\frac{d}{dx}\left(b\right), dove a=y e b=\mathrm{tanh}\left(x\right)^2\ln\left(\mathrm{sech}\left(y\right)\right). Applicare la formula: \frac{d}{dx}\left(x\right)=1. Applicare la formula: \frac{d}{dx}\left(ab\right)=\frac{d}{dx}\left(a\right)b+a\frac{d}{dx}\left(b\right), dove d/dx=\frac{d}{dx}, ab=\mathrm{tanh}\left(x\right)^2\ln\left(\mathrm{sech}\left(y\right)\right), a=\ln\left(\mathrm{sech}\left(y\right)\right), b=\mathrm{tanh}\left(x\right)^2 e d/dx?ab=\frac{d}{dx}\left(\mathrm{tanh}\left(x\right)^2\ln\left(\mathrm{sech}\left(y\right)\right)\right). Applicare la formula: \frac{d}{dx}\left(x^a\right)=ax^{\left(a-1\right)}\frac{d}{dx}\left(x\right), dove a=2 e x=\mathrm{tanh}\left(x\right).
d/dx(y=ln(sech(y))tanh(x)^2)
Risposta finale al problema
$y^{\prime}=\frac{2\mathrm{sech}\left(x\right)^2\ln\left(\mathrm{sech}\left(y\right)\right)\mathrm{tanh}\left(x\right)}{1+\mathrm{tanh}\left(x\right)^2\mathrm{tanh}\left(y\right)}$