Question

Obtenha a derivada de cada uma das seguintes funções:
a) f(x) = (x2 + 1) • tg x
b) f(x)= x²+3x+1/ x-2

Answer 1

Resposta:

Explicação passo a passo:

Sejam:

1) y = u.v ⇒ y' = u.v' + vu'

$2)y=\frac{u}{v}\implies y'=\frac{vu'-uv'}{v^2} \\\\3)y=a^{u} \implies y'=a^{u}*lna*u'$

$a)f(x)=(x^2+1)*tgx\\\\f'(x)=(x^2+1)*(tgx)'+tgx*(x^2+1)'\\\\f'(x)= (x^2+1)sec^2x+tgx(2x+0)\\\\f'(x)=(x^2+1)sec^2x+2x*tgx\\\\\\b)f(x)=\frac{x^2+3x+1}{x-2} \\\\f'(x)=\frac{(x-2)*(x^2+3x+1)'-(x^2+3x+1)(x-2)'}{(x-2)^2} \\\\f'(x)=\frac{(x-2)(2x+3+0)-(x^2+3x+1)*(1+0)}{(x-2)^2} \\\\f'(x)=\frac{(x-2)(2x+3)-(x^2+3x+1)}{(x-2)^2} \\\\f'(x)=\frac{2x^2+3x-4x-6-x^2-3x-1}{(x-2)^2} \\\\f'(x)=\frac{x^2-4x-7}{(x-2)^2}\\\\\\4)f(x)= e^{x^{2}+5x } \\\\f'(x)= e^{x^{2}+5x }*lne*(x^2+5x)'\\\\f'(x)= e^{x^{2}+5x } *1*(2x+5)$

$f'(x)=(2x+5)*e^{x^{2} +5x} \\\\f'(-1)=[2(-1)+5]*e^{(-1)^{2} +5(-1)} \\\\f'(-1)=(-2+5)*e^{1-5}\\\\f'(-1)=3e^-^4\\\\f'(-1)=\frac{3}{e^4} \\\\m=\frac{3}{e^4}$

m é o coeficiente angular no ponto x = -1