Question

Qual seria a Derivada da função v(t) = (2t+9)²(8-t)³ ?​

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Regra~da~Cadeia}\\\sf (f[g(x)])'=f'[g(x)]\cdot g'(x)\\\underline{\rm Derivada~da~pot\hat encia}\\\sf[x^n]'=nx^{n-1}\\\underline{\rm Derivada~do~produto}\\\sf [f(x)\cdot g(x)]'=f'(x)\cdot g(x)+f(x)\cdot g'(x)\end{array}}$

$\small\boxed{\begin{array}{l}\sf v(t)=(2t+9)^2(8-t)^3\\\sf v'(t)\\\sf =2\cdot(2t+9)\cdot(2t+9)'\cdot (8-t)^3+(2t+9)^2\cdot3(8-t)^2\cdot(8-t)'\\\sf v'(t)\\\sf=2\cdot(2t+9)\cdot2\cdot(8-t)^3+(2t+9)^2\cdot3(8-t)^2\cdot(-1)\\\sf v'(t)\\\sf=4\cdot(2t+9)(8-t)^3-3\cdot(2t+9)^2\cdot(8-t)^2\\\sf colocando~2t+9~e~(8-t)^2~em\,evid\hat encia~temos:\\\sf v'(t)=(2t+9)(8-t)^2[4(8-t)-3(2t+9)]\\\sf v'(t)=(2t+9)(8-t)^2[32-4t-6t-27]\end{array}}$

$\Large\boxed{\begin{array}{l}\sf v'(t)=(2t+9)(8-t)^2\cdot(5-10t)\\\large\boxed{\boxed{\boxed{\boxed{\sf\blue{v'(t)=5(2t+9)(8-t)^2(1-2t)}}}}}\red{\checkmark}\end{array}}$