What is the derivative for $y = tan (x) + x^{\frac{5}{2}} + 2 x$ $y=tan(x)+x^(5/2)+2x$ ?

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What is the derivative for $y = tan (x) + x^{\frac{5}{2}} + 2 x$ $y=tan(x)+x^(5/2)+2x$ ?

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Answer 1 · 2018-05-02T15:31:04

$\frac{d y}{d x} = {sec}^{2} x + \frac{5}{2} x^{\frac{3}{2}} + 2$ $dy/dx=sec^2x+5/2x^(3/2)+2$

Explanation:

$y = tan x + x^{\frac{5}{2}} + 2 x$ $y=tanx+x^(5/2)+2x$

$\frac{d}{d x} tan u = {sec}^{2} u \frac{d u}{d x}$ $color(green)(d/dxtanu=sec^2ucolor(blue)((du)/dx$ $\to Where u is a function of x$ $color(red)(rarr"Where u is a function of " x$

$\frac{d}{d x} u^{n} = n \cdot u^{n - 1} \frac{d u}{d x}$ $color(green)(d/dxu^n=n*u^(n-1)color(blue)((du)/dx$ $\to Where u is a function of x$ $color(red)(rarr"Where u is a function of " x$

$\frac{d y}{d x} = {sec}^{2} x \times 1 + \frac{5}{2} x^{\frac{3}{2}} \times 1 + 2 \times 1$ $dy/dx=color(green)(sec^2x)color(blue)(xx1)+color(green)(5/2x^(3/2))color(blue)(xx1)+color(green)(2color(blue)(xx1)$

$= {sec}^{2} x + \frac{5}{2} x^{\frac{3}{2}} + 2$ $=sec^2x+5/2x^(3/2)+2$

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What is the derivative for $y = tan (x) + x^{\frac{5}{2}} + 2 x$ $y=tan(x)+x^(5/2)+2x$ ?

1 Answer

Explanation:

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