What is the implicit derivative of #25=sin(xy)/x-3xy#?

Question

Ezekiel Alexander · Accepted Answer

Long time since tried this but I am having a go!
#color(blue)((dy)/dx =(sin(xy)-yxcos(xy))/(x^2cos(xy)-3x^3)+(3y)/(cos(xy)-3x))#
#color(red)("You will need to check this!!")# #d/(dx)( (sin(xy))/x -3xy) =d/dx(25)# #d/dx((sin(xy))/x) - d/dx(3xy)=0# Using std forms #d/dx(uv) =v(du)/dx + u(dv)/dx # #color(white)(xxxxxxxxxx)d/dx(u/v) = (v(du)/dx - u(dv)/dx )/v^2# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Let #color(white)(xx)w =uv -> xy# then # color(green)((dw)/dx ->(dy)/dx= y +x(dy)/dx ...........................(1))_"confirmed"# Let #t=sin(xy) = sin(w) # then #color(green)( (dt)/dx->(dy)/dx = (y+x(dy)/dx)cos(xy) .............(2))_"confirmed"# #color(green)(d/dx(3xy) = 3(y+x(dy)/dx) "from (1) "................(3))_"confirmed"# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Consider: #d/dx((sin(xy))/x) -> u/v# #(x{ycos(xy)+x(dy/dx)cos(xy)} - sin(xy))/x^2 ......._color(red)("corrected y-> ycos(xy)")# #color(green)((ycos(xy))/x +(dy)/dx cos(xy) -sin(xy)/x^2.........(4))_color(red)(("corrected" y->ycos(xy)))# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Combining (3) and (4)")# #{y/x cos(xy)+(dy)/dx cos(xy)-sin(xy)/x^2}-{3y+3x(dy)/dx}=0_("corrected " y->ycos(xy))# Collecting like terms #color(red)("(rebuilt calculations)")# #(dy)/dx cos(xy) -3x(dy)/dx color(red)( =) (sin(xy))/x^2-y/xcos(xy)+3y_(color(red)(" corrected " y->ycos(xy)))# #color(white)(x)# #(dy)/dx color(red)( =) 1/(cos(xy)-3x) ((sin(xy))/x^2 -y/xcos(xy)+3y)# #(dy)/dxcolor(red)( =) 1/(cos(xy)-3x)((sin(xy)-yxcos(xy))/x^2+3y)# #color(blue)((dy)/dx =(sin(xy)-yxcos(xy))/(x^2cos(xy)-3x^3)+(3y)/(cos(xy)-3x))#

Luke Alvarez · Answer

By working step-wise taking derivatives of individual components you should end up with
#(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)# This is complex so check carefully before assuming what follows is valid: Breaking the equation up into small pieces and working on the derivatives for each: Part 1: The #color(red)(xy)# component of #25=sin(color(red)(xy))/x-3xy# Finding the derivative of #xy# is of the form for finding the derivative of #uv# and we know #(d(uv))/(dx) = v(du)/(dx)+u(dv)/(dx)# so #color(white)("XXX")(d(xy))/(dx) = y(cancel(dx))/(cancel(dx)) + x(dy)/(dx)# #color(white)("XXXXXX")=y+x(dy)/(dx)# Part 2: The #color(red)(sin(xy))# component of #25=color(red)(sin(xy))/x-3xy# Finding the derivative of #sin(xy)# is of the form for finding the derivative of #f(g(u))# and we know #color(white)("XXX")(df(g(x)))/(dx)=(df(g(x)))/(d(g(x))) * (dg(x))/(dx)# with #f(x) = sin(x)# and #g(x)=xy# so #color(white)("XXX")(dsin(xy))/(dx) = cos(xy)*(y+x(dy)/(dx))# Part 3: The #color(red)(sin(xy)/x)# component of #25= color(red)(sin(xy)/x)-3xy# Finding the derivative of #sin(xy)/x# is of the form for finding the derivative of #u/v# and we know #color(white)("XXX")d(u/v)/dx = (v(du)/(dx) -u(dv)/(dx))/(v^2)# with #v=x# and #u=sin(xy)# so #color(white)("XXX")(d(sin(xy))/x)/(dx)= (x*(cos(xy)*(y+x(dy)/(dx)))-sin(xy)(cancel(dx))/(cancel(dx)))/(x^2)# #color(white)("XXXXXXXX")=(cos(xy)(y+x(dy)/(dx)))/x - (sin(xy))/(x^2)# Part 4: The #color(red)(xy)# component of #25 = sin(xy)/x - 3color(red)(xy)# from Part 1 we already have #color(white)("XXX")#(d(xy))/(dx) = y+x(dy)/(dx)# Putting the pieces together Take the derivative of both sides #d(25)/(dx) = (d(sin(x)/x-3xy))/dx# #0 = (dsin(x)/x)/dx-3((d(xy))/(dx))# #(cos(xy)(y+x(dy)/(dx)))/x - (sin(xy))/(x^2)-3(y+x(dy)/(dx))=0# #(ycos(xy))/x +cos(xy)(dy)/(dx)-sin(xy)/(x^2)-3y-3x(dy)/(dx)=0# #(ycos(xy))/x-sin(xy)/(x^2)-3y = (3x-cos(xy))(dy)/(dx)# #(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)#

Christopher Agresta · Answer

undefined To find the implicit derivative of $ 25 = \frac{\sin(xy)}{x} - 3xy $, differentiate both sides of the equation with respect to $ x $ using the product rule and chain rule. The implicit derivative will be:

$$ \frac{d}{dx} (25) = \frac{d}{dx} \left(\frac{\sin(xy)}{x}\right) - \frac{d}{dx} (3xy) $$

$$ 0 = \frac{y\cos(xy)}{x} - \frac{\sin(xy)}{x^2} - 3y - 3x\frac{dy}{dx} $$

$$ \frac{dy}{dx} = \frac{\sin(xy)}{x^2} - \frac{y\cos(xy)}{x} - 3y $$