What is the derivative for #y=tan(x)+x^(5/2)+2x#?

1 Answer
Yahia M.
May 2, 2018

#dy/dx=sec^2x+5/2x^(3/2)+2#

Explanation:

#y=tanx+x^(5/2)+2x#

#color(green)(d/dxtanu=sec^2ucolor(blue)((du)/dx##color(red)(rarr"Where u is a function of " x#

#color(green)(d/dxu^n=n*u^(n-1)color(blue)((du)/dx##color(red)(rarr"Where u is a function of " x#

#dy/dx=color(green)(sec^2x)color(blue)(xx1)+color(green)(5/2x^(3/2))color(blue)(xx1)+color(green)(2color(blue)(xx1)#

#=sec^2x+5/2x^(3/2)+2#

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