Differentiate the following implicit dunction and find dy/dx ln (3xy)=2xy^2
I have now shown you how to do several of these kind of derivatives.
It is time for you to show us some of your own efforts.
What have you got so far ?
ln (3x) + ln (y) = 2xy^2
Is it need to separate in the bracket?
How to separate x and y for RHS
There is really no need to have brackets on the RS
but I would mentally look at it as
(2x)(y^2) so the product rule is more obvious
so .....
3/(3x) + (dy/dx)/y = (2x)(2y dy/dx) + (y^2)(2)
multiply each term by y
y/x + dy/dx = 4xy^2 dy/dx + 2y^3
dy/dx - 4xy^2 dy/dx = 2y^3 - y/x
dy/dx (1 - 4xy^2) = 2y^3 - y/x
dy/dx = (2y^3 - y/x)/(1 - 4xy^2)
= (2xy^3 - y)/(x - 4x^2 y^2)
again , check my steps
To differentiate the given implicit function and find dy/dx, we can follow these steps:
Step 1: Rewrite the given implicit function in an explicit form:
Start with the given implicit function: ln(3xy) = 2xy^2
To make it explicit, exponentiate both sides using the base e (natural logarithm) because ln and e are inverse functions:
e^(ln(3xy)) = e^(2xy^2)
3xy = e^(2xy^2)
Step 2: Differentiate both sides of the equation with respect to x:
To differentiate the left side, we need to apply the product rule, where u = 3x and v = y:
(d/dx)(u * v) = u * (d/dx)v + v * (d/dx)u
Differentiating the right side is a little more complex. We have e^(2xy^2), which is an exponential function with respect to x and a function of x and y. To differentiate this kind of function, we need to apply the chain rule:
(d/dx)(e^(2xy^2)) = d/dx (f(g(x))) = f'(g(x)) * g'(x)
In this case, let f(u) = e^u and g(x) = 2xy^2.
So, f'(u) = e^u (the derivative of e^u is e^u itself).
And, g'(x) = 2y * 1 (since the derivative of 2xy^2 with respect to x is 2y due to the chain rule).
Let's proceed with the differentiation:
Left side: d/dx (3xy) = 3y + 3x * (d/dx)y
Right side: d/dx (e^(2xy^2)) = e^(2xy^2) * (2y)
Step 3: Substitute the derivatives back into the equation:
3y + 3x * (d/dx)y = e^(2xy^2) * (2y)
Step 4: Solve for (d/dx)y:
Rearrange the equation to isolate (d/dx)y:
3x * (d/dx)y = e^(2xy^2) * (2y) - 3y
(d/dx)y = (e^(2xy^2) * (2y) - 3y) / (3x)
Hence, we have found the derivative of y with respect to x, (dy/dx), for the given implicit function ln(3xy) = 2xy^2:
(dy/dx) = (e^(2xy^2) * (2y) - 3y) / (3x)