Question

(20) 2. find the following derivatives (no need to simplify).
$\frac{d}{dx}(7x^{3}+x^{pi}-12x + sqrt{x}-5)=$
$\frac{d}{du}(\frac{1}{u}+\frac{3}{u^{5}}+sqrt5{u^{3}}-4e^{3})=$
$\frac{d}{dt}(t^{3}+9e^{t}-1)(t^{7}-t + 3t^{-4})=$
$\frac{d}{dx}(\frac{7 + x-x^{3}}{1+sqrt{x}-x^{2}})=$

Explanation:

Step1: Apply sum - difference rule

$\frac{d}{dx}(7x^{3}+x^{\pi}-12x + \sqrt{x}-5)=\frac{d}{dx}(7x^{3})+\frac{d}{dx}(x^{\pi})-\frac{d}{dx}(12x)+\frac{d}{dx}(\sqrt{x})-\frac{d}{dx}(5)$

Step2: Use power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$

$=7\times3x^{2}+\pi x^{\pi - 1}-12\times1+\frac{1}{2}x^{-\frac{1}{2}}-0=21x^{2}+\pi x^{\pi - 1}-12+\frac{1}{2\sqrt{x}}$

Step3: For $\frac{d}{du}(\frac{1}{u}+\frac{3}{u^{5}}+\sqrt[5]{u^{3}}-4e^{3})$, rewrite terms

$\frac{d}{du}(u^{-1}+3u^{-5}+u^{\frac{3}{5}}-4e^{3})$

Step4: Apply sum - difference and power rules

$=-u^{-2}+3\times(- 5)u^{-6}+\frac{3}{5}u^{-\frac{2}{5}}-0=-\frac{1}{u^{2}}-\frac{15}{u^{6}}+\frac{3}{5u^{\frac{2}{5}}}$

Step5: For $\frac{d}{dt}(t^{3}+9e^{t}-1)(t^{7}-t + 3t^{-4})$, use product rule $(uv)' = u'v+uv'$

Let $u=t^{3}+9e^{t}-1$, $u'=3t^{2}+9e^{t}$; $v=t^{7}-t + 3t^{-4}$, $v'=7t^{6}-1-12t^{-5}$
Then $\frac{d}{dt}(t^{3}+9e^{t}-1)(t^{7}-t + 3t^{-4})=(3t^{2}+9e^{t})(t^{7}-t + 3t^{-4})+(t^{3}+9e^{t}-1)(7t^{6}-1-12t^{-5})$

Step6: For $\frac{d}{dx}(\frac{7 + x-x^{3}}{1+\sqrt{x}-x^{2}})$, use quotient rule $(\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}$

Let $u = 7 + x-x^{3}$, $u'=1-3x^{2}$; $v=1+\sqrt{x}-x^{2}=1+x^{\frac{1}{2}}-x^{2}$, $v'=\frac{1}{2}x^{-\frac{1}{2}}-2x$
Then $\frac{d}{dx}(\frac{7 + x-x^{3}}{1+\sqrt{x}-x^{2}})=\frac{(1 - 3x^{2})(1+\sqrt{x}-x^{2})-(7 + x-x^{3})(\frac{1}{2\sqrt{x}}-2x)}{(1+\sqrt{x}-x^{2})^{2}}$

Answer:

1. $21x^{2}+\pi x^{\pi - 1}-12+\frac{1}{2\sqrt{x}}$
2. $-\frac{1}{u^{2}}-\frac{15}{u^{6}}+\frac{3}{5u^{\frac{2}{5}}}$
3. $(3t^{2}+9e^{t})(t^{7}-t + 3t^{-4})+(t^{3}+9e^{t}-1)(7t^{6}-1-12t^{-5})$
4. $\frac{(1 - 3x^{2})(1+\sqrt{x}-x^{2})-(7 + x-x^{3})(\frac{1}{2\sqrt{x}}-2x)}{(1+\sqrt{x}-x^{2})^{2}}$