Question

Calcular la derivada:
y= e^2x / x^1/2
y= log2 (x^3+2x)
y= ln (1/x)
y= log ((1+x)/(1-x))^1/2

Answer 1

$y=\frac{e^{2x}}{x^{\frac{1}{2}}} \\ \\ y'=\frac{x^{\frac{1}{2}}[\frac{d}{dx}(e^{2x})]-[e^{2x}(\frac{d}{dx}(x^{\frac{1}{2}}))]}{(x^{\frac{1}{2}})^{2}}=\frac{x^{\frac{1}{2}}(e^{2x}2)-[e^{2x}(\frac{1}{2\sqrt{x}})]}{x} \\ \\ y'=\frac{e^{2x}2x^{\frac{1}{2}}-\frac{e^{2x}}{2\sqrt{x}}}{x}=\frac{\frac{4e^{2x}x-e^{2x}}{2\sqrt{x}}}{x} \\ \\ y'=\frac{4e^{2x}x-e^{2x}}{2x^{\frac{3}{2}}}$

$y=\log_2\left(x^3+2x\right) \\ \\ y'=\frac{d}{dx}[log_2(x^{3}+2x)][\frac{d}{dx}(x^{3}+2x)]=\frac{1}{(x^{3}+2x)ln(2)}(3x^{2}+2) \\ \\ y'=\frac{3x^{2}+2}{(x^{3}+2x)ln(2)}$

$y=ln\frac{1}{x} \\ \\ y'=\frac{\frac{d}{dx}(\frac{1}{x})}{\frac{1}{x}}=\frac{-\frac{1}{x^{2}}}{\frac{1}{x}}=-\frac{x}{x^{2}} \\ \\ y'=-\frac{1}{x}$

$y=log(\frac{1+x}{1-x})^{\frac{1}{2}} \\ \\ y'=\frac{d}{dx}[log_{10}(\frac{1+x}{1-x})^{\frac{1}{2}}][\frac{d}{dx}(\frac{1+x}{1-x})] \\ \\ y'=\frac{1}{2\sqrt{log_{10}(\frac{1+x}{1-x})}}(\frac{2}{ln(10)(1+x)(1-x)}) \\ \\ y'=\frac{1}{ln(10)\sqrt{log_{10}(\frac{1+x}{1-x})}(1+x)(1-x)}$

Espero haberte ayudado. ¡Saludos!