Question

Encontre as derivadas das seguintes funções:
d) h(p)= p³/3 + 3/p³
e) g(t)= (2t^4 -1)(5t³+6t)
usando limites

Answer 1

Oi 

Só consegui por derivadas mesmo. :( 

$h(p)= \frac{p^3}{3} + \frac{3}{p^3} \\ \\ h'(p)= 3\frac{p^2}{3} + 3p^{-3} \\ \\ h'(p)=p^2-3.3p^{-4} \\ \\ h'(p)=p^2- \frac{9}{p^{4}}$

$g(t)=(2t^4-1)(5t^3+6t) \\ \\ g(t)=10t^7+12t^5-5t^3-6t \\ \\ g'(t)=7.10t^6+5.12t^4-3.5t^2-6 \\ \\ g'(t)=70t^6+60t^4-15t^2-6$

Answer 2

Cálculo de derivadas de funções pela definição:

$\boxed{\begin{array}{c}f'(x)=\lim\limits_{s\to 0}\dfrac{f(x+s)-f(x)}{s}\end{array}}$

(onde esse limite existir e for finito)


d) $h(p)=\dfrac{p^3}{3}+\dfrac{3}{p^3}$

$\displaystyle h'(p)=\lim_{s\to 0}\,\frac{h(p+s)-h(p)}{s}\\\\\\ =\lim_{s\to 0}\frac{\left[\frac{(p+s)^3}{3}+\frac{3}{(p+s)^3}\right]-\left[\frac{p^3}{3}+\frac{3}{p^3}\right]}{s}\\\\\\ =\lim_{s\to 0}\frac{\frac{(p+s)^3}{3}+\frac{3}{(p+s)^3}-\frac{p^3}{3}-\frac{3}{p^3} }{s}\\\\\\ =\lim_{s\to 0}\frac{\frac{(p+s)^3}{3}-\frac{p^3}{3}+\frac{3}{(p+s)^3}-\frac{3}{p^3}}{s}\\\\\\ =\lim_{s\to 0}\left[\frac{(p+s)^3-p^3}{3}+\frac{3p^3-3(p+s)^3}{(p+s)^3\cdot p^3} \right]\cdot \frac{1}{s}$

$\displaystyle =\lim_{s\to 0}\frac{\left[(p+s)^3-p^3\right]\cdot (p+s)^3\cdot p^3+3\cdot\left[3p^3-3(p+s)^3\right]}{3\cdot(p+s)^3\cdot p^3\cdot s}\\\\\\ =\lim_{s\to 0}\frac{\left[(p+s)^3-p^3 \right ]\cdot(p+s)^3\cdot p^3-9\cdot\left[(p+s)^3-p^3\right]}{3\cdot (p+s)^3\cdot p^3\cdot s}\\\\\\ =\lim_{s\to 0}\frac{\left[(p+s)^3-p^3\right]\cdot \left[(p+s)^3\cdot p^3-9\right]}{3\cdot (p+s)^3\cdot p^3\cdot s}$

$\displaystyle =\lim_{s\to 0}\frac{\left[(\diagup\!\!\!\! p^3+3p^2 s+3ps^2+s^3)-\diagup\!\!\!\! p^3\right]\cdot \left[(p+s)^3\cdot p^3-9 \right ]}{3\cdot(p+s)^3\cdot p^3\cdot s}\\\\\\ =\lim_{s\to 0}\frac{(3p^2 s+3ps^2+s^3)\cdot \left[(p+s)^3\cdot p^3-9 \right ]}{3\cdot(p+s)^3\cdot p^3\cdot s}\\\\\\ =\lim_{s\to 0}\frac{\diagup\!\!\!\! s\cdot (3p^2+3ps+s^2)\cdot \left[(p+s)^3\cdot p^3-9 \right ]}{3\cdot(p+s)^3\cdot p^3\cdot \diagup\!\!\!\! s}\\\\\\ =\lim_{s\to 0}\frac{(3p^2+3ps+s^2)\cdot \left[(p+s)^3\cdot p^3-9 \right ]}{3\cdot(p+s)^3\cdot p^3}$

$\displaystyle =\frac{(3p^{2}+3p\cdot 0+0^{2})\cdot [(p+0)^3\cdot p^3-9]}{3\cdot \left(p+0 \right )^{3}\cdot p^{3}}\\\\\\ =\frac{3p^2\cdot (p^6-9)}{3p^6}\\\\\\ =\frac{3p^2\cdot (p^4\cdot p^2-9)}{3p^2\cdot p^4}\\\\\\ =\frac{3p^2\cdot p^4\left(p^2-\frac{9}{p^4}\right)}{3p^6}\\\\\\ =\frac{3p^6\cdot \left(p^2-\frac{9}{p^4}\right)}{3p^6}\\\\\\ \therefore~~\boxed{\begin{array}{c}h'(p)=p^2-\dfrac{9}{p^4} \end{array}}$

_________

e) $g(t)=(2t^4-1)(5t^3+6t)$

$\displaystyle g'(t)=\lim_{s\to 0}\frac{g(t+s)-g(t)}{s}\\\\\\ =\lim_{s\to 0}\frac{\big(2(t+s)^4-1\big)\big(5(t+s)^3+6(t+s)\big)-(2t^4-1)(5t^3+6t)}{s}\\\\\\ =\lim_{s\to 0}\left[\big(2(t+s)^4-1\big)\big(5(t+s)^3+6(t+s)\big)-(2t^4-1)(5t^3+6t)\right]\cdot \dfrac{1}{s}$


Ao numerador, subtraímos e somamos $(2t^4-1)\big(5(t+s)^3+6(t+s)\big):$

$\begin{array}{ll} =\lim\limits_{s\to 0}&\left[\big(2(t+s)^4-1\big)\big(5(t+s)^3+6(t+s)\big)-(2t^4-1)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.(2t^4-1)\big(5(t+s)^3+6(t+s)\big)-(2t^4-1)(5t^3+6t)\right]\cdot\dfrac{1}{s}\\\\\\ =\lim\limits_{s\to 0}&\left[\big(2(t+s)^4-1-(2t^4-1)\big)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.(2t^4-1)\big(5(t+s)^3+6(t+s)-(5t^3+6t)\big)\right]\cdot\dfrac{1}{s}\\\\\\ =\lim\limits_{s\to 0}&\left[\big(2(t^4+4t^3s+6t^2s^2+4ts^3+s^4)-1-2t^4+1\big)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.(2t^4-1)\big(5(t^3+3t^2s+3ts^2+s^3)+6(t+s)-5t^3-6t\big)\right]\cdot\dfrac{1}{s} \end{array}$

$\begin{array}{ll} =\lim\limits_{s\to 0}&\left[\big(2t^4+8t^3s+12t^2s^2+8ts^3+2s^4-1-2t^4+1\big)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.(2t^4-1)\big(5t^3+15t^2s+15ts^2+5s^3+6t+6s-5t^3-6t\big)\right]\cdot\dfrac{1}{s}\\\\\\ =\lim\limits_{s\to 0}&\left[\big(8t^3s+12t^2s^2+8ts^3+2s^4\big)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.(2t^4-1)\big(15t^2s+15ts^2+5s^3+6s\big)\right]\cdot\dfrac{1}{s}\\\\\\ =\lim\limits_{s\to 0}&\left[s\cdot \big(8t^3+12t^2s+8ts^2+2s^3\big)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.s\cdot (2t^4-1)\big(15t^2+15ts+5s^2+6\big)\right]\cdot\dfrac{1}{s}\\\\\\ =\lim\limits_{s\to 0}&\diagup\!\!\!\! s\cdot \left[\big(8t^3+12t^2s+8ts^2+2s^3\big)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.(2t^4-1)\big(15t^2+15ts+5s^2+6\big)\right]\cdot\dfrac{1}{\diagup\!\!\!\! s} \end{array}$

$\begin{array}{ll} =\lim\limits_{s\to 0}&\left[\big(8t^3+12t^2s+8ts^2+2s^3\big)\big(5(t+s)^3+6(t+s)\big)\right.\\\\ &+\left.(2t^4-1)\big(15t^2+15ts+5s^2+6\big)\right]\\\\\\ =&\left[\big(8t^3+12t^2\cdot 0+8t\cdot 0^2+2\cdot 0^3\big)\big(5(t+0)^3+6(t+0)\big)\right.\\\\ &+\left.(2t^4-1)\big(15t^2+15t\cdot 0+5\cdot 0^2+6\big)\right] \end{array}\\\\\\\\ \therefore~~\boxed{\begin{array}{c}g'(t)=8t^3\cdot \big(5t^3+6t\big)+(2t^4-1)\cdot \big(15t^2+6\big)\end{array}}$


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Bons estudos! :-)