Question

derive a função
a)x^2+4x+3/raiz de x

Answer 1

Note que essa função não é diferenciável em $x=0$, pois ela não é contínua em $x=0$. Então, para $x\neq0$, podemos escrever

$f(x)=\dfrac{x^{2}+4x+3}{\sqrt{x}}=\dfrac{x^{2}+4x+3}{x^{1/2}}\\\\\\f(x)=\dfrac{x^{2}}{x^{1/2}}+\dfrac{4x^{1}}{x^{1/2}}+\dfrac{3}{x^{1/2}}\\\\\\f(x)=x^{2-(1/2)}+4x^{1-(1/2)}+3x^{-1/2}\\\\\f(x)=x^{3/2}+4x^{1/2}+3x^{-1/2}$

Derivando $f$:

$\dfrac{d}{dx}f(x)=\dfrac{d}{dx}(x^{3/2})+\dfrac{d}{dx}(4x^{1/2})+\dfrac{d}{dx}(3x^{-1/2})\\\\\\f'(x)=\dfrac{3}{2}\,x^{(3/2)-1}+4\dfrac{d}{dx}(x^{1/2})+3\dfrac{d}{dx}(x^{-1/2})\\\\\\f'(x)=\dfrac{3}{2}\,x^{1/2}+4\cdot\dfrac{1}{2}\,x^{(1/2)-1}+3\bigg(-\dfrac{1}{2}\bigg)x^{-(1/2)-1}\\\\\\\boxed{\boxed{f'(x)=\dfrac{3}{2}x^{1/2}+2x^{-1/2}-\dfrac{3}{2}x^{-3/2}}}$

Podemos simplificar a expressão de $f'$. Como o domínio de $f$ é o conjunto dos reais positivos, temos:

$f'(x)=\dfrac{3}{2}x^{1/2}+\dfrac{2}{x^{1/2}}-\dfrac{3}{2}\dfrac{1}{x^{3/2}}\\\\\\f'(x)=\dfrac{3x^{4/2}}{2x^{3/2}}+\dfrac{4x^{2/2}}{2x^{3/2}}-\dfrac{3}{2x^{3/2}}\\\\\\f'(x)=\dfrac{3x^{2}+4x^{1}-3}{2x^{3/2}}\\\\\\\boxed{\boxed{f'(x)=\dfrac{3x^{2}+4x-3}{2\sqrt{x^{3}}}}}$

Answer 2

Resposta:

bouua

Explicação passo a passo:

mais ou menos