Question

Calcule o coeficiente angular da reta tangente ao gráfico da função f(x)= e^x²+ 5x no ponto de abscissa —1.

Answer 1

$\Large\boxed{\begin{array}{l}\rm f(x)=e^{x^2+5x}\\\rm f'(x)=e^{x^2+5x}\cdot(2x+5)\\\rm f'(-1)=e^{(-1)^2+5\cdot(-1)}\cdot(2\cdot(-1)+5)\\\rm f'(-1)=e^{1-5}\cdot(-2+5)\\\rm f'(-1)=e^{-4}\cdot3\\\rm f'(-1)=\dfrac{3}{e^4}\checkmark\end{array}}$