Encontre f^{\prime} e f^{\prime \prime} :
f(x)=x^{3 / 2}\left(x+c e^{x}\right)

Passo 1

O enunciado pede para calcular as derivadas primeira e segunda de
f(x)=x^{3/2}\bigl(x + c e^x\bigr).
O conteúdo principal que vamos usar é a regra do produto e a regra da cadeia para potências e exponenciais.

Passo 2

Para encontrar f'(x), escrevemos
f(x)=u(x)\,v(x),
com
u(x)=x^{3/2}, \qquad v(x)=x + c e^x.
Então
u'(x)=\frac{3}{2}x^{1/2}, \qquad v'(x)=1 + c e^x.
Pela regra do produto,
f'(x)=u'(x)\,v(x)+u(x)\,v'(x) =\frac{3}{2}x^{1/2}\bigl(x+c e^x\bigr) +x^{3/2}\bigl(1+c e^x\bigr).

Passo 3

Podemos agrupar f'(x) como
f'(x) =x^{1/2}\Bigl[\frac{3}{2}\bigl(x+c e^x\bigr)+x\bigl(1+c e^x\bigr)\Bigr] =x^{1/2}\Bigl[\frac{3}{2}x+\frac{3c}{2}e^x+x+c x e^x\Bigr]
\boxed{f'(x) =x^{1/2}\Bigl(\frac{5}{2}x+c e^x\bigl(\tfrac{3}{2}+x\bigr)\Bigr).}

Passo 4

Para f''(x), vamos derivar
f'(x) =\frac{3}{2}x^{1/2}\bigl(x+c e^x\bigr) +x^{3/2}\bigl(1+c e^x\bigr).
Separe em dois termos:
T_1=\tfrac{3}{2}x^{1/2}(x+c e^x) e T_2=x^{3/2}(1+c e^x).
— Derivando T_1:
T_1=\frac{3}{2}x^{3/2}+\frac{3c}{2}x^{1/2}e^x
T_1'=\frac{3}{2}\cdot\frac{3}{2}x^{1/2} +\frac{3c}{2}\Bigl(\tfrac12x^{-1/2}e^x+x^{1/2}e^x\Bigr) =\frac{9}{4}x^{1/2}+\frac{3c}{4}x^{-1/2}e^x+\frac{3c}{2}x^{1/2}e^x.
— Derivando T_2:
T_2=x^{3/2}+c\,x^{3/2}e^x
T_2'=\frac{3}{2}x^{1/2} +c\Bigl(\frac{3}{2}x^{1/2}e^x+x^{3/2}e^x\Bigr) =\frac{3}{2}x^{1/2}+\frac{3c}{2}x^{1/2}e^x+c\,x^{3/2}e^x.

Passo 5

Somando T_1' e T_2':
f''(x) =\Bigl(\tfrac{9}{4}x^{1/2}+\tfrac{3}{2}x^{1/2}\Bigr) +\frac{3c}{4}x^{-1/2}e^x +\Bigl(\tfrac{3c}{2}x^{1/2}e^x+\tfrac{3c}{2}x^{1/2}e^x\Bigr) +c\,x^{3/2}e^x
=\frac{15}{4}x^{1/2} +\frac{3c}{4}x^{-1/2}e^x +3c\,x^{1/2}e^x +c\,x^{3/2}e^x.

Resposta

\boxed{% f'(x)=x^{1/2}\Bigl(\tfrac{5}{2}x+c e^x(\tfrac{3}{2}+x)\Bigr), \quad f''(x)=\frac{15}{4}x^{1/2} +\frac{3c}{4}x^{-1/2}e^x +3c\,x^{1/2}e^x +c\,x^{3/2}e^x. }

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