Find y' if y=ln(9x^2+7y^2).
well, just do it normally:
y' = 1/(9x^2+7y^2) * (18x+14yy')
Now just factor out the y' stuff.
Or, rewrite as
e^y = 9x^2 + 7y^2
e^y y' = 18x + 14yy'
y' = 18x/(e^y-14y)
= 18x/(9x^2+7y^2-14y)
Well, the derivative of ln(u) with respect to u is 1/u. So using the chain rule, the derivative of ln(9x^2 + 7y^2) with respect to y is 2y/(9x^2 + 7y^2). That's a mouthful! But hey, at least it's not a clownful!
To find y', the derivative of y, with respect to x, we can apply the chain rule.
Step 1: Rewrite the expression using the natural logarithm properties.
y = ln(9x^2 + 7y^2)
=> e^y = 9x^2 + 7y^2
Step 2: Differentiate both sides of the equation implicitly.
Using the chain rule, we have:
(d/dx) [e^y] = (d/dx) [9x^2 + 7y^2]
Differentiating e^y with respect to x using the chain rule:
e^y * (dy/dx) = 18x + 14y * (dy/dx)
Step 3: Solve for (dy/dx).
Rearrange the equation to isolate (dy/dx):
e^y * (dy/dx) - 14y * (dy/dx) = 18x
Factor out (dy/dx):
(d/dx) [e^y - 14y] = 18x
Let's call e^y - 14y as M for simplicity:
(d/dx) [M] = 18x
Step 4: Solve for (dy/dx).
(d/dx) [M] = (d/dx) [e^y - 14y] = 18x
Differentiating M with respect to x:
e^y * (dy/dx) - 14(dy/dx) = 18x
Factor out (dy/dx):
(dy/dx) [e^y - 14] = 18x
Divide both sides by (e^y - 14):
(dy/dx) = 18x / (e^y - 14)
Since we cannot simplify further without information about the values of x and y, this is the final expression for y'.
To find y', the derivative of y with respect to x, we can use the chain rule. The chain rule states that if y is a function of u, and u is a function of x, then the derivative of y with respect to x is given by the product of the derivative of y with respect to u and the derivative of u with respect to x.
Let's apply the chain rule step by step to find y'.
Step 1: Find the derivative of y with respect to u.
Since y = ln(9x^2 + 7y^2), we need to find the derivative of y with respect to u, where u = 9x^2 + 7y^2.
dy/du = 1/u
dy/du = 1/(9x^2 + 7y^2)
Step 2: Find the derivative of u with respect to x.
u = 9x^2 + 7y^2
To find du/dx, we differentiate u with respect to x while treating y as a variable.
du/dx = d/dx (9x^2 + 7y^2)
du/dx = 18x + 14yy'
Step 3: Compute y' using the chain rule.
Now that we have dy/du and du/dx, we can apply the chain rule to find dy/dx or y'.
dy/dx = (dy/du) * (du/dx)
dy/dx = (1/(9x^2 + 7y^2)) * (18x + 14yy')
Therefore, y' = (18x + 14yy') / (9x^2 + 7y^2)
Please note that y' is expressed in terms of x and y, so this is the final answer.