Differentiate: $f ( x ) = \left( x ^ { 2 } + 1 \right) ( x + 3 )$
A) 2x + 1
B) 6x + 1
C) $3 x ^ { 2 } + 6 x + 1$
D) $x ^ { 2 } + 1$

Question

Quizplus · Accepted Answer

To differentiate $f(x) = (x^2 + 1)(x + 3)$, we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$. Let $u(x) = x^2 + 1$ and $v(x) = x + 3$. Then, $u'(x) = 2x$ and $v'(x) = 1$. Applying the product rule, we get $f'(x) = (2x)(x + 3) + (x^2 + 1)(1) = 2x^2 + 6x + x^2 + 1 = 3x^2 + 6x + 1$.

Differentiate f(x)=(x2+1)(x+3)f ( x ) = \left( x ^ { 2 } + 1 \right) ( x + 3 )f(x)=(x2+1)(x+3)

Differentiate $f ( x ) = \left( x ^ { 2 } + 1 \right) ( x + 3 )$