Use implicit differentiation to find dy/dx:
a) xrooty-ye^x=1
b) sinxy + ln y=x
x^2+3xy+2y^2=7
Y : x^2 -3xy +2y^2 =0 and hence show that the ratio x : y is 1:1 or 2:1
Sure! Let's find the derivatives of the functions using implicit differentiation.
a) To find dy/dx for the equation x√y - ye^x = 1, we treat y as a function of x and differentiate both sides of the equation with respect to x.
Taking the derivative of x√y with respect to x, we have:
d/dx(x√y) = d/dx(1 + ye^x)
To simplify, we can use the chain rule, where d/dx(f(g(x))) = f'(g(x)) * g'(x).
Thus, we get:
√y + x * 1/(2√y) * dy/dx = ye^x
Next, we differentiate -ye^x with respect to x:
d/dx(-ye^x) = d/dx(1)
Using the chain rule, we get:
-1 * e^x * dy/dx - y * e^x = 0
Now, rearranging the equation, we have:
√y + xy'/(2√y) = ye^x + y'e^x
Simplifying, we get:
y'/(2√y) - y' * e^x = -√y + ye^x
To find dy/dx, let's isolate y' on one side of the equation:
y'/(2√y) - y' * e^x = -√y + ye^x
y'/(2√y) - y' * e^x = -√y + ye^x
Common denominator:
y'/(2√y) - y' * e^x = -2y/(2√y) + 2ye^x/(2√y)
Combine like terms:
y' * (1/(2√y) - e^x) = -2y/(2√y) + 2ye^x/(2√y)
Simplify:
y' * (1/(2√y) - e^x) = (-2y + 2ye^x)/(2√y)
Finally, we can solve for y':
y' = (-2y + 2ye^x)/(2√y - 2ye^x)
This is the expression for dy/dx for the equation x√y - ye^x = 1.
b) For the equation sin(xy) + ln(y) = x, let's differentiate both sides of the equation with respect to x:
Take the derivative of sin(xy) with respect to x:
d/dx(sin(xy)) = d/dx(x)
Using the chain rule, we get:
(y * cos(xy)) * dy/dx + x = 1
Next, differentiate ln(y) with respect to x:
d/dx(ln(y)) = d/dx(x)
Using the chain rule, we get:
1/y * dy/dx + 0 = 1
Now, rearranging the equation, we have:
(y * cos(xy)) * dy/dx + x = 1 - 1/y * dy/dx
To find dy/dx, let's isolate it on one side of the equation:
(y * cos(xy)) * dy/dx + 1/y * dy/dx = 1 - x
Common denominator:
(y * cos(xy) + 1) * dy/dx = 1 - x
Finally, we can solve for dy/dx:
dy/dx = (1 - x)/(y * cos(xy) + 1)
This is the expression for dy/dx for the equation sin(xy) + ln(y) = x.