Answer to Question #262721 in Calculus for Alunsina

Question

Find y'. Simplify all answers.

a. x²y + xy² - y³ + x² = 2

b. sqrt(2xy) + y²-xy²=1

c. sinx + cosy = sinxcosy

d. y = x^x^x

e. y = (sinx)^2x

f. ln(x²+y²) + 2arctan(y/x) = 1

g. (x+y)³ = (x-y)²

Answer 1

a.

$(x^2y + xy^2- y^3+ x^2 )' = (2)'$

$2xy+x^2y'+y^2+2xyy'-3y^2y'+2x=0$

$y'=\dfrac{2xy+y^2+2x}{3y^2-2xy-x^2}$

b.

$(\sqrt{2xy} + y^2-xy^2)'=(1)'$

$\sqrt{\dfrac{y}{2x}}+\sqrt{\dfrac{x}{2y}}y'+2yy'-y^2-2xyy'=0$

$y'=\dfrac{y^2-\sqrt{\dfrac{y}{2x}}}{\sqrt{\dfrac{x}{2y}}+2y-2xy}$

c.

$(\sin x + \cos y )'= (\sin x\cos y)'$

$\cos x-y'\sin y=\cos x\cos y-y'\sin x \sin y$

$y'=\dfrac{\cos x\cos y-\cos x}{\sin x \sin y-\sin y}$

d.

$y = x^{x^x}$

$\ln y=x^x\ln x$

$\dfrac{\ln y}{\ln x}=x^x$

$\ln(\dfrac{\ln y}{\ln x})=x\ln x$

$(\ln(\ln y)-\ln(\ln x))'=(x\ln x)'$

$\dfrac{1}{y\ln y}y'-\dfrac{1}{x\ln x}=\ln x+1$

$y'=y\ln y(\dfrac{1}{x\ln x}+\ln x+1)$

$y'=x^{x^x}x^x\ln x(\dfrac{1}{x\ln x}+\ln x+1)$

e.

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

$(\ln y)'=(2x\ln(\sin x))'$

$\dfrac{y'}{y}=2\ln(\sin x)+2x(\dfrac{\cos x}{\sin x})$

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

f)

$(\ln(x²+y²) + 2\arctan(y/x))' = (1)'$

$\dfrac{2x+2yy'}{x^2+y^2}+\dfrac{2}{1+y^2/x^2}(\dfrac{y'}{x}-\dfrac{y}{x^2})=0$

$x+yy'+xy'-y=0$

$y'=\dfrac{y-x}{y+x}$

g)

$((x+y)^3)' =( (x-y)^2)'$

$3(x+y)^2(1+y')=2(x-y)(1-y')$

$y'=\dfrac{2(x-y)-3(x+y)^2}{3(x+y)^2+2(x-y)}$