Question

Найдите производную y=(2x+3)^2-4tgx, при x0=0

Answer 1

$y(x)=(2x+3)^2-4tgx\\y`(x_0)-?\\x_0=0\\\\y`(x)=3(2x+3)(2x+3)`- \frac{4}{cos^2x}=3(2x+3)*2- \frac{4}{cos^2x}=\\\\=6(2x+3)-\frac{4}{cos^2x} \\\\y`(x_0)=y`(0)=6(2*0+3)-\frac{4}{cos^20}=6*3- \frac{4}{1^2}=18-4=14\\\\y`(x_0)=14$