If dy/dx = x²y², then d²y/dx² = ?
Well, if dy/dx = x²y², then d²y/dx² must be double the trouble, or as I like to call it, "twice the spice!" So, d²y/dx² = 2x³y²! It's like taking the original equation and giving it a big ol' comedy makeover!
To find the second derivative, d²y/dx², you'll need to differentiate the equation dy/dx = x²y² with respect to x again.
Starting with the given equation: dy/dx = x²y²
Differentiating both sides with respect to x, we use the product rule (d(uv)/dx = u * dv/dx + v * du/dx) and chain rule (d(uv)/dx = (du/dx)v + (dv/dx)u) to obtain:
d²y/dx² = (d/dx)(x²y²)
= d/dx(x²) * y² + x² * d/dx(y²)
= 2x * y² + x² * (d/dx(y²))
Next, we need to differentiate (d/dx)(y²). Using the chain rule, we have:
(d/dx)(y²) = (d/dy)(y²) * (dy/dx)
Since dy/dx = x²y², we can substitute this back into the equation:
(d/dx)(y²) = (d/dy)(y²) * (dy/dx)
= 2y * (x²y²)
= 2x²y³
Now, we substitute this result back into our original equation:
d²y/dx² = 2x * y² + x² * (d/dx)(y²)
= 2x * y² + x² * (2x²y³)
= 2x * y² + 2x^4y³
Therefore, the second derivative d²y/dx² is equal to 2x * y² + 2x^4y³.
To find the second derivative of y with respect to x (∂²y/∂x²), we need to differentiate the given expression dy/dx = x²y² again.
First, let's rewrite the given expression as follows:
dy/dx = x² * y²
To differentiate this expression, we will use the chain rule. According to the chain rule, if we have a function of the form f(g(x)), the derivative of f(g(x)) with respect to x is df/dx = df/dg * dg/dx.
In our case, f(g(x)) = g(x)², and g(x) = xy. So, by applying the chain rule, we have:
d²y/dx² = d/dx (dy/dx)
= d/dx (x² * y²)
= d/dx (x²) * y² + x² * d/dx (y²)
= 2x * y² + x² * d/dx (y²)
Now, d/dx (y²) can be found using the chain rule again. Let's denote h(z) = z², and z = y(x). So, d/dx (y²) = d/dx (h(g(x)) = dh/dg * dg/dx.
We know that dh/dz = 2z and dg/dx = y'. Substituting these values, we get:
d/dx (y²) = dh/dz * dg/dx
= 2y * y'
= 2yy'
Finally, substituting this value back into our previous expression, we have:
d²y/dx² = 2x * y² + x² * (2yy')
= 2xy² + 2x²yy'
= 2xy² + 2x²y * dy/dx
Therefore, the second derivative of y with respect to x (∂²y/∂x²) is 2xy² + 2x²y * dy/dx, which can also be written as 2xy² + 2x²y³.
f(x,y) = x^2 y^2
df(x,y)/dx = x^2 (2 y dy/dx) + y^2(2 x dx/dx)
but dy/dx = x^2y^2
so
x^2(2 y x^2y^2) + 2 x y^2
2 x^4 y^3 + 2 x y^2
2 x y^2 (x^2 y + 1)