if y=tan u , u=v-1÷v, and v=ln x , what is the value of dy÷dx at x=e?
if y=tan u , u=v-1÷v, and v=ln x , what is the value of dy÷dx at x=e?
To find the value of dy/dx at x=e, we need to differentiate y with respect to x and substitute x=e into the derivative expression.
Let's start with finding the derivative of y = tan(u) using the chain rule.
First, compute the derivative of u with respect to x:
du/dx = (d/dx)(v - (1/v))
Using the power rule and the chain rule:
du/dx = (d/dx)(v) - (d/dx)(1/v)
du/dx = (dv/dx) - (-1/v^2)(dv/dx)
du/dx = (dv/dx) + (dv/dx)/v^2
du/dx = (2dv/dx)/v^2
Now, compute the derivative of y with respect to u:
dy/du = d(tan(u))/du
dy/du = sec^2(u)
Next, use the chain rule to find dy/dx in terms of v and x:
dy/dx = (dy/du) * (du/dx)
dy/dx = [(2dv/dx)/v^2] * sec^2(u)
Let's compute the values of u and v at x=e to substitute them into the expression for dy/dx:
Given that v = ln(x), we can substitute x=e to find v:
v = ln(e) = 1
Given that u = v - (1/v), we can substitute v=1:
u = 1 - (1/1) = 1 - 1 = 0
Now, substitute u=0 and v=1 into the expression for dy/dx:
dy/dx = [(2dv/dx)/v^2] * sec^2(u)
dy/dx = [(2dv/dx)/1] * sec^2(0)
dy/dx = 2dv/dx
Therefore, the value of dy/dx at x=e is simply equal to 2 times the derivative of v with respect to x evaluated at x=e.
However, since we don't have any additional information about v, we cannot determine the specific value of dy/dx at x=e without knowing the function v or any additional conditions.
To find the value of dy/dx at x=e, we need to differentiate the given expression with respect to x and substitute x=e.
Let's start by finding dy/du and du/dv:
1. Given y = tan(u), we can differentiate both sides with respect to u:
dy/du = d/dy(tan(u))
= sec^2(u) * du/dy
= sec^2(u)
2. Next, we have u = (v-1)/v. Let's differentiate both sides with respect to v:
du/dv = d/dv((v-1)/v)
= (v * (1) - (v-1) * (1))/v^2
= 1/v^2
Finally, we need to find dv/dx and substitute x=e:
3. Given v = ln(x), we differentiate both sides with respect to x:
dv/dx = d/dx(ln(x))
= 1/x
Substituting x=e into dv/dx, we get:
dv/dx = 1/e
Now, we can find dy/dx using the chain rule:
dy/dx = dy/du * du/dv * dv/dx
Substituting the values we found earlier:
dy/dx = sec^2(u) * (1/v^2) * (1/e)
To find the value of dy/dx at x=e, we need to substitute the values u and v at x=e:
Since v = ln(x):
v = ln(e) = 1
And since u = (v-1)/v:
u = (1-1)/1 = 0
Now we can substitute u=0 and v=1 into dy/dx:
dy/dx = sec^2(0) * (1/1^2) * (1/e)
= 1 * 1 * (1/e)
= 1/e
Therefore, the value of dy/dx at x=e is 1/e.
why post it twice???
Assuming you meant u=(v-1)/v, we have
dy'dx = dy/du * du/dv * dv/dx
= (sec^2 u)(1/v^2)(1/x)
at x=e, v=1, u=0, y=0
dy/dx = (1)(1)(1/e) = 1/e