Q.Using the product rule, differentiate $y = (3x+1)(x^2-2)$.

Apply the power rule term by term: the derivative is $20x^3-9x^2+4x-11$.

Using $\dfrac{d}{dx}(x^n)=nx^{n-1}$ on $5x^4-3x^3+2x^2-11x+7$:

$\frac{d}{dx}(5x^4)=20x^3,\quad \frac{d}{dx}(-3x^3)=-9x^2,$

$\frac{d}{dx}(2x^2)=4x,\quad \frac{d}{dx}(-11x)=-11,\quad \frac{d}{dx}(7)=0.$

Adding: $20x^3-9x^2+4x-11$.

Product rule: $\dfrac{d}{dx}(uv)=u\,v'+v\,u'$.

If $u=u(x)$ and $v=v(x)$ are two differentiable functions, the derivative of their product is

$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}.$

This is the product rule of differentiation.

Using the power rule, $\dfrac{d}{dx}(5x^6)=30x^5$.

Step 1 — the power rule states $\dfrac{d}{dx}(x^n)=nx^{n-1}$.

Step 2 — constants multiply through (constant-multiple rule):

$\dfrac{d}{dx}(5x^6)=5\cdot\dfrac{d}{dx}(x^6)=5\cdot 6x^{6-1}=30x^{5}$

The derivative is $6r + 4$.

Using the power rule $\dfrac{d}{dr}(r^n) = n r^{n-1}$ and the fact that the derivative of a constant is $0$, differentiate each term:

$\frac{d}{dr}(3r^2) = 6r, \qquad \frac{d}{dr}(4r) = 4, \qquad \frac{d}{dr}(7) = 0.$

Adding,

$\frac{d}{dr}(3r^2 + 4r + 7) = 6r + 4.$

Correct option: (d) $\dfrac{1}{2\sqrt{x}}$.

Express the root as a power and use $\dfrac{d}{dx}x^n = n x^{n-1}$:

$\sqrt{x} = x^{1/2}, \qquad \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.$