derivative of ((Sqrt(x+3y)+sqrt(3xy))=(sqrt(21)+sqrt(90)) at point (6,5)google
Check that (6,5) lies on the curve by substituting x=6,y=5 and see if the equality holds. (yes, it holds).
Differentiate:
((Sqrt(x+3y)+sqrt(3xy))=(sqrt(21)+sqrt(90))
(1/2)/sqrt(x+3y)*(1+3y')+(1/2)/sqrt(3xy)*(3y+3xy')=0
transpose and cross multiply:
-sqrt(x+3y)*(1+3y')=sqrt(3xy)*(3y+3xy')
Solve for y'
y'(x,y)=-(sqrt(3*y+x)+3^(3/2)*y*sqrt(x*y))/(3*sqrt(3*y+x)+3^(3/2)*x*sqrt(x*y))
so
y'(6,5)=(-5*3^(3/2)*sqrt(30)-sqrt(21))/(2*3^(5/2)*sqrt(30)+3*sqrt(21))
=-0.796....
please check my arithmetic.
To find the derivative of the given function, you can use the rules of differentiation. In this case, there are two terms: √(x + 3y) and √(3xy).
Let's calculate the derivative step by step:
Step 1: Differentiate the first term with respect to x and the second term with respect to x separately.
For the first term, √(x + 3y):
To differentiate this term, we need to use the chain rule. Let's define u = x + 3y.
So, √(x + 3y) = √u.
Now, differentiate √u with respect to x:
(d/dx) √u = (d/du) √u * (du/dx) = (1/2√u) * (d/dx)(x + 3y) = (1/2√u) * (1 + 0) = 1 / (2√(x + 3y)).
For the second term, √(3xy):
We can apply the chain rule in a similar manner:
Let's define v = 3xy.
So, √(3xy) = √v.
Differentiating √v with respect to x:
(d/dx) √v = (d/dv) √v * (dv/dx) = (1/2√v) * (d/dx)(3xy) = (1/2√v) * (3y) = (3y)/(2√(3xy)).
Step 2: Combine the derivatives of both terms to obtain the derivative of the given equation.
(d/dx) (√(x + 3y) + √(3xy)) = 1 / (2√(x + 3y)) + (3y)/(2√(3xy)).
Step 3: Evaluate the derivative at the given point (6, 5).
Substitute the values of x = 6 and y = 5 into the derivative expression:
(1 / (2√(6 + 3(5)))) + (3(5))/(2√(3(6)(5))).
Simplifying further, we get:
(1 / (2√(6 + 15))) + (15)/(2√(3(6)(5))).
Now, to find the numerical value of the derivative at this point, you can use Google or any calculator to evaluate the expression:
(1 / (2√(6 + 15))) + (15)/(2√(3(6)(5))).
This will give you the derivative of the function at the point (6, 5).