question 1. find the derivatives of the following functions using the product rule. (a) $h(x)=(x^{3}-3x)(2x…

Question 1(a)

Explanation:

Step1: Recall product - rule

The product - rule states that if $h(x)=u(x)v(x)$, then $h^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)$. Let $u(x)=x^{3}-3x$ and $v(x)=2x - 6$.

Step2: Differentiate $u(x)$

$u^\prime(x)=\frac{d}{dx}(x^{3}-3x)=3x^{2}-3$

Step3: Differentiate $v(x)$

$v^\prime(x)=\frac{d}{dx}(2x - 6)=2$

Step4: Apply product - rule

$h^\prime(x)=(3x^{2}-3)(2x - 6)+(x^{3}-3x)\times2$ $=6x^{3}-18x^{2}-6x + 18+2x^{3}-6x$ $=8x^{3}-18x^{2}-12x + 18$

Answer:

$h^\prime(x)=8x^{3}-18x^{2}-12x + 18$

Question 1(b)

Explanation:

Step1: Recall product - rule

The product - rule states that if $s(t)=u(t)v(t)$, then $s^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t)$. Let $u(t)=38\sqrt{t}=38t^{\frac{1}{2}}$ and $v(t)=2t^{3}-t^{2}+7t - 18$.

Step2: Differentiate $u(t)$

$u^\prime(t)=\frac{d}{dt}(38t^{\frac{1}{2}})=38\times\frac{1}{2}t^{-\frac{1}{2}} = 19t^{-\frac{1}{2}}$

Step3: Differentiate $v(t)$

$v^\prime(t)=\frac{d}{dt}(2t^{3}-t^{2}+7t - 18)=6t^{2}-2t + 7$

Step4: Apply product - rule

$s^\prime(t)=19t^{-\frac{1}{2}}(2t^{3}-t^{2}+7t - 18)+38t^{\frac{1}{2}}(6t^{2}-2t + 7)$ $=38t^{\frac{5}{2}}-19t^{\frac{3}{2}}+133t^{\frac{1}{2}}-342t^{-\frac{1}{2}}+228t^{\frac{5}{2}}-76t^{\frac{3}{2}}+266t^{\frac{1}{2}}$ $=266t^{\frac{5}{2}}-95t^{\frac{3}{2}}+399t^{\frac{1}{2}}-342t^{-\frac{1}{2}}$

Answer:

$s^\prime(t)=266t^{\frac{5}{2}}-95t^{\frac{3}{2}}+399t^{\frac{1}{2}}-342t^{-\frac{1}{2}}$

Question 2(a)

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $r(x)=\frac{u(x)}{v(x)}$, then $r^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^{2}}$. Let $u(x)=3x^{2}+5x$ and $v(x)=x + 4$.

Step2: Differentiate $u(x)$

$u^\prime(x)=\frac{d}{dx}(3x^{2}+5x)=6x + 5$

Step3: Differentiate $v(x)$

$v^\prime(x)=\frac{d}{dx}(x + 4)=1$

Step4: Apply quotient - rule

$r^\prime(x)=\frac{(6x + 5)(x + 4)-(3x^{2}+5x)\times1}{(x + 4)^{2}}$ $=\frac{6x^{2}+24x+5x + 20-3x^{2}-5x}{(x + 4)^{2}}$ $=\frac{3x^{2}+24x + 20}{(x + 4)^{2}}$

Answer:

$r^\prime(x)=\frac{3x^{2}+24x + 20}{(x + 4)^{2}}$

Question 2(b)

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $s(t)=\frac{u(t)}{v(t)}$, then $s^\prime(t)=\frac{u^\prime(t)v(t)-u(t)v^\prime(t)}{v(t)^{2}}$. Let $u(t)=3t^{4}+2$ and $v(t)=t^{3}+1$.

Step2: Differentiate $u(t)$

$u^\prime(t)=\frac{d}{dt}(3t^{4}+2)=12t^{3}$

Step3: Differentiate $v(t)$

$v^\prime(t)=\frac{d}{dt}(t^{3}+1)=3t^{2}$

Step4: Apply quotient - rule

$s^\prime(t)=\frac{12t^{3}(t^{3}+1)-(3t^{4}+2)\times3t^{2}}{(t^{3}+1)^{2}}$ $=\frac{12t^{6}+12t^{3}-9t^{6}-6t^{2}}{(t^{3}+1)^{2}}$ $=\frac{3t^{6}+12t^{3}-6t^{2}}{(t^{3}+1)^{2}}$

Answer:

$s^\prime(t)=\frac{3t^{6}+12t^{3}-6t^{2}}{(t^{3}+1)^{2}}$