Question

f(x)= 7/2x^2+1/6x^3+x+3/4 derivada​

Answer 1

Resposta:

a)Vamos usar a regra do produto:

y=f(x)·g(x) ⇒ y'=f(x)g'(x) + g(x)f'(x)

\begin{gathered}f(x) = (7x^2+6x)^7 (3x-1)^4 \\ f'(x) = (7x^2+6x)^7 4(3x-1)^3 +(3x-1)^47(7x^2+6x)^6\end{gathered}

f(x)=(7x

2

+6x)

7

(3x−1)

4

f

′

(x)=(7x

2

+6x)

7

4(3x−1)

3

+(3x−1)

4

7(7x

2

+6x)

6

b)Vamos usar a regra da potencia

Se y=a^ny=a

n

então y'=n a^{n-1}y

′

=na

n−1

\begin{gathered}f(x) = (7x+ \frac{x^2}{2} +3)^3 \\ f'(x)=3(7x+\frac{x^2}{2}+3)^2\end{gathered}

f(x)=(7x+

2

x

2

+3)

3

f

′

(x)=3(7x+

2

x

2

+3)

2

c)Regra da potencia

\begin{gathered}f(x) = \sqrt{(3x^2+6x-2)^2} \\ f(x)=[(3x^2+6x-2)^2]^ \frac{1}{2} \\ f(x)=3x^2+6x-2 \\ f'(x)=6x+6\end{gathered}

f(x)=

(3x

2

+6x−2)

2

f(x)=[(3x

2

+6x−2)

2

]

2

1

f(x)=3x

2

+6x−2

f

′

(x)=6x+6