Find the slope of the tangent line to the curve √(1x+2y) + √(1xy) = 8.24 at the point (2,8)?
I know you have to use implicit differentiation, but the radicals keep making me mess up algebraically. Is the changing the radicals to exponents the fastest way? please show me the steps thanks
(1x)^(1/2) + (2y)^(1/2) + (1xy)^(1/2)=8.24
Before I attempt this .....
I am curious why you would put a coefficient of 1 in front of the variables, such as in
(1x + 2y) and (1xy)
At this stage of Calculus, they would certainly be understood and not needed.
I am tempted to guess you meant
(1/x + 2y) and √(1/xy)
please clarify.
no it's actually 1x and 1xy. It's not division.
√(1x+2y) + √(1xy) = 8.24
changing the radicals to exponents is really the only way, explicitly or implicitly, but
√(1x+2y) ≠ √(1x) + √(2y) !!!
Anyway, we have, using implicit differentiation (and deleting those useless 1 coefficients),
√(x+2y) + √(xy) = 8.24
1/2√(x+2y) (1+2y') + 1/2√(xy) (y+xy') = 0
1/2√(x+2y) + y/2√(xy) + y'/√(x+2y) + xy'/2√(xy) = 0
y'(1/√(x+2y) + x/2√(xy)) = -(1/2√(x+2y) + y/2√(xy))
y' =
-(y√(x+2y) + √(xy)) / (x√(x+2y) + 2√(xy))
thanks
To find the slope of the tangent line to the given curve at the point (2,8), we can use implicit differentiation. However, as you mentioned, the presence of square roots can make the algebraic manipulation more complicated. One way to simplify the equation and facilitate differentiation is to change the radicals to exponents, as you suggested.
Let's start by rewriting the equation in exponential form:
√(1x+2y) + √(1xy) = 8.24
can be rewritten as:
(1x+2y)^(1/2) + (1xy)^(1/2) = 8.24
Now, to differentiate with respect to x, we treat y as a function of x using the chain rule. Let's denote the square root of a function as f(x):
f(x) = (1x+2y)^(1/2) + (1xy)^(1/2)
Differentiating both sides with respect to x, we get:
(d/dx) f(x) = (d/dx) ((1x+2y)^(1/2) + (1xy)^(1/2))
Using the chain rule, the derivative of the first term is:
(d/dx) f(x) = (1/2) * (1x+2y)^(-1/2) * (d/dx) (1x+2y) + (d/dx) ((1xy)^(1/2))
The derivative of (1x+2y) with respect to x is simply 1 since y is a function of x. Therefore, the first term becomes:
(d/dx) f(x) = (1/2) * (1x+2y)^(-1/2) * 1 + (d/dx) ((1xy)^(1/2))
For the second term, we apply the chain rule again:
(d/dx) ((1xy)^(1/2)) = (1/2) * ((1xy)^(-1/2)) * (d/dx) (1xy)
The derivative of (1xy) with respect to x becomes:
(d/dx) ((1xy)^(1/2)) = (1/2) * ((1xy)^(-1/2)) * (y + x(dy/dx))
Substituting this result back into the first term, we have:
(d/dx) f(x) = (1/2) * (1x+2y)^(-1/2) + (1/2) * ((1xy)^(-1/2)) * (y + x(dy/dx))
Now, to find the slope of the tangent line at the point (2,8), we substitute x = 2 and y = 8 into the derivative equation:
(d/dx) f(2) = (1/2) * (1(2)+2(8))^(-1/2) + (1/2) * ((1(2)(8))^(-1/2)) * (8 + 2(dy/dx))
Simplifying this equation will give you the slope of the tangent line at the point (2,8).