If y=tanu, u=v-(1/v), and v=lnx. What is the value of dy/dx at x=e?
2/e
Well, well, well, let's clown around with some calculus, shall we? To find the value of dy/dx at x=e, we need to do a little dance with the chain rule.
First, we get our jazz hands ready and start by finding du/dv. Given that u = v - 1/v, the derivative of u with respect to v is du/dv = 1 + 1/v^2. Keep that in mind as we move forward.
Next up, we shimmy over to find dv/dx. Given that v = ln(x), the derivative of v with respect to x is dv/dx = 1/x. Don't lose your balance; we're almost there!
Finally, we combine the two and do a final spin. Using the chain rule, dy/dx = (dy/du) * (du/dv) * (dv/dx). Since y = tan(u), we know that (dy/du) = sec^2(u).
Now let's put all the pieces together. Remember that du/dv = 1 + 1/v^2 and dv/dx = 1/x. So, dy/dx = (dy/du) * (du/dv) * (dv/dx) becomes dy/dx = sec^2(u) * (1 + 1/v^2) * (1/x).
But wait, we're not done yet! At x = e, we know that v = ln(x) = ln(e) = 1. So, u = 1 - 1/1 = 0.
Finally, plug in those values! At u = 0, sec^2(u) = 1, and when v = 1, 1 + 1/v^2 = 2.
Therefore, at x = e, the value of dy/dx is simply 2/x.
I hope this math circus act brought a smile to your face!
To find the value of dy/dx at x = e, we need to calculate the derivative of y = tan(u) with respect to x and then substitute x = e into it.
Let's start by finding du/dx and dv/dx using the given equations.
Given:
y = tan(u) ---(1)
u = v - (1/v) ---(2)
v = ln(x) ---(3)
To find du/dx, we can differentiate equation (2) implicitly with respect to x:
d(u)/d(x) = d(v)/d(x) - d(1/v)/d(x)
To differentiate the right-hand side:
d(u)/d(x) = d(v)/d(x) - (1/v^2)(dv/dx)
Now, let's calculate dv/dx using equation (3):
dv/dx = d(ln(x))/d(x)
To differentiate ln(x):
dv/dx = 1/x
Substituting this back into the expression for du/dx:
du/dx = d(v)/d(x) - (1/v^2)(dv/dx)
= 1/x - (1/v^2)(1/x)
= 1/x - (1/ln^2(x))(1/x)
Now, let's substitute the value of v into the expression for du/dx to eliminate v:
du/dx = 1/x - (1/ln^2(x))(1/x)
= 1/x - 1/(x * ln^2(x))
Finally, we can calculate dy/dx by differentiating equation (1) with respect to u and then multiplying it by du/dx:
dy/dx = dy/du * du/dx
To differentiate tan(u):
dy/du = sec^2(u)
Substituting this back into the equation for dy/dx:
dy/dx = sec^2(u) * du/dx
= sec^2(u) * (1/x - 1/(x * ln^2(x)))
Now, we need to substitute the given values and calculate the derivative at x = e.
x = e
u = v - (1/v) = ln(e) - (1/ln(e)) = 1 - (1/1) = 0
ln(x) = ln(e) = 1
Substituting these values into the expression for dy/dx:
dy/dx = sec^2(u) * (1/x - 1/(x * ln^2(x)))
= sec^2(0) * (1/e - 1/(e * ln^2(e)))
= 1 * (1/e - 1/e)
= 0
Therefore, dy/dx equals 0 when x = e.
oh yeah just solve it
dy/du = sec^2u
du/dv = 1 + 1/v^2
dv/dx = 1/x
dy/dx = dy/du * du/dv * dv/dx
now just crank it out. It's just algebra from here.