Answer to Question #277301 in Calculus for KING

Question

1. Find the derivative of the following functions:

a. 𝑓(𝑥) = tan^-1x+cotx / 5cscx

 b. 𝑦=3 −ln(2𝑥+1)

c. 𝑔(𝑥)=𝑥𝑒^x−sin𝑥ln(5𝑥)

Answer 1

a.

$(tan^{-1}x)'=\frac{1}{1+x^2}$

$cscx=1/sinx,cotx=cosx/snx$

$(cotx / 5cscx)'=\frac{(cotx)'cscx-(cscx)'cotx}{5csc^2x}$

$(cotx)'=-csc^2x=-1/sin^2x$

$(cscx)'=-cscxcotx=-cosx/sin^2x$

$(cotx / 5cscx)'=\frac{(-1/sin^2x)/sinx+(cosx/sin^2x)cosx/sinx}{5/sin^2x}=$

$=\frac{(cos^2x-1)/sin^3x}{5/sin^2x}=-\frac{sinx}{5}$

$f'(x)=\frac{1}{1+x^2}-sinx/5$

b.

$y'=-2/(2x+1)$

c.

$g'(x)=e^x+xe^x-cosxln(5x)-sinx/x$

$(tan^{-1}x)'=\frac{1}{1+x^2}$

$cscx=1/sinx,cotx=cosx/snx$

$(cotx / 5cscx)'=\frac{(cotx)'cscx-(cscx)'cotx}{5csc^2x}$

$(cotx)'=-csc^2x=-1/sin^2x$

$(cscx)'=-cscxcotx=-cosx/sin^2x$

$(cotx / 5cscx)'=\frac{(-1/sin^2x)/sinx+(cosx/sin^2x)cosx/sinx}{5/sin^2x}=$

$=\frac{(cos^2x-1)/sin^3x}{5/sin^2x}=-\frac{sinx}{5}$