Apply Product rule
to find the derivative
a
b
c

Ask by Norris Stuart. in Mexico

Mar 20,2025

Question

Apply Product rule to find the derivative a $$ f(x)=x(x-1) $$ b $$ f(x)=2 x(x+1) $$ c $$ f(x)=x^{2} \sqrt{x+1} $$

UpStudy AI · Accepted Answer

**a) Derivada de $$ f(x)=x(x-1) $$**

Definimos:
- $$ u(x)=x $$ con $$ u'(x)=1 $$
- $$ v(x)=x-1 $$ con $$ v'(x)=1 $$

Aplicando la regla del producto:

$$
f'(x)=u'(x)v(x)+u(x)v'(x)=1\cdot (x-1)+x\cdot 1=(x-1)+x=2x-1
$$

---

**b) Derivada de $$ f(x)=2x(x+1) $$**

Definimos:
- $$ u(x)=2x $$ con $$ u'(x)=2 $$
- $$ v(x)=x+1 $$ con $$ v'(x)=1 $$

Aplicando la regla del producto:

$$
f'(x)=u'(x)v(x)+u(x)v'(x)=2\cdot (x+1)+2x\cdot 1=2(x+1)+2x=2x+2+2x=4x+2
$$

---

**c) Derivada de $$ f(x)=x^{2}\sqrt{x+1} $$**

Notamos que $$ \sqrt{x+1}=(x+1)^{\frac{1}{2}} $$.

Definimos:
- $$ u(x)=x^2 $$ con $$ u'(x)=2x $$
- $$ v(x)=(x+1)^{\frac{1}{2}} $$ y usando la regla de la potencia, 
$$
v'(x)=\frac{1}{2}(x+1)^{-\frac{1}{2}}
$$

Aplicando la regla del producto:

$$
\begin{aligned}
f'(x) &= u'(x)v(x)+u(x)v'(x) \
      &= 2x\,(x+1)^{\frac{1}{2}}+x^2\left(\frac{1}{2}(x+1)^{-\frac{1}{2}}\right) \
      &= 2x\sqrt{x+1}+\frac{x^2}{2\sqrt{x+1}}
\end{aligned}
$$

Cada derivada ha sido calculada aplicando la regla del producto paso a paso.
**a) Derivada de $$ f(x)=x(x-1) $$:** $$ f'(x) = 2x - 1 $$ **b) Derivada de $$ f(x)=2x(x+1) $$:** $$ f'(x) = 4x + 2 $$ **c) Derivada de $$ f(x)=x^{2}\sqrt{x+1} $$:** $$ f'(x) = 2x\sqrt{x+1} + \frac{x^2}{2\sqrt{x+1}} $$

Apply Product rule
to find the derivative
a
b
c

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