Answer to Question #92551 in Calculus for joseph Ko

Question

1. y = (x+1) (×-2) / (x+1) (x+2)
2. ln xy = 2 raised to x+y
3. y= csc( 2 raised to log with base of 2 x squared)
4. y= arctan ( e raised to x squared 3 raised to x cube)

Answer 1

Compute a derivative of y with respect to x using initial formulas.

1. $y=\frac{(x+1)(x-2)}{(x+1)(x+2)}$

$y=\frac{x-2}{x+2}$

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

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

$y'=\frac{4}{(x+2)^2}$

Answer: $\frac{4}{(x+2)^2}$

2. $\ln{(xy)}=2^{x+y}$

$(\ln{(xy)})'=(2^{x+y})'$

$\frac{1}{xy}(xy)'=2^{x+y}\ln{2}\ (x+y)'$

$\frac{y+xy'}{xy}=2^{x+y}\ln{2}\ (1+y')$

$\frac{1}{x}+\frac{y'}{y}=2^{x+y}\ln{2}+\ 2^{x+y}\ln{2}\ y'$

$\frac{y'}{y}-2^{x+y}\ln{2}\ y'=2^{x+y}\ln{2}-\frac{1}{x}$

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

$y'=\frac{2^{x+y}\ln{2}-\frac{1}{x}}{\frac{1}{y}-2^{x+y}\ln{2}}$

$y'=-\frac{2^{x+y}\ln{2}-\frac{1}{x}}{2^{x+y}\ln{2}-\frac{1}{y}}$

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

Answer:$-\frac{y(2^{x+y}x\ln{2}-1)}{x(2^{x+y}y\ln{2}-1)}$

3. $y=\cosec{2^{log_2{x^2}}}$

$y=\cosec{x^2}$

$y=\frac{1}{\sin{x^2}}$

$y'=-\frac{(\sin{x^2})'}{\sin^2{x^2}}$

$y'=-\frac{2x\cdot \cos{x^2}}{\sin^2{x^2}}$

Answer:$-\frac{2x\cdot \cos{x^2}}{\sin^2{x^2}}$

4. $y=\arctan{(e^{x^2}3^{x^3})}$

$y'=\frac{(e^{x^2}3^{x^3})'}{1+(e^{x^2}3^{x^3})^{2}}$

$y'=\frac{(e^{x^2})'3^{x^3}+(3^{x^3})'e^{x^2}}{1+(e^{x^2}3^{x^3})^{2}}$

$y'=\frac{2x\cdot e^{x^2}3^{x^3}+3x^2\cdot3^{x^3}(\ln{3})\cdot e^{x^2}}{1+(e^{x^2}3^{x^3})^{2}}$

$y'=\frac{x\cdot e^{x^2}3^{x^3}(3x\ln{3}+2)}{e^{2x^2}3^{2x^3}+1}$

Answer:$\frac{x\cdot e^{x^2}3^{x^3}(3x\ln{3}+2)}{e^{2x^2}3^{2x^3}+1}$