Find dy/dx for x=1/t+1 and y=1/t-1
To find dy/dx, we can first find dy/dt and dx/dt, and then use the chain rule.
To find dy/dt, we can differentiate the equation y = 1/t - 1 with respect to t:
dy/dt = d/dt(1/t - 1)
= -1/t^2
To find dx/dt, we can differentiate the equation x = 1/(t+1) with respect to t:
dx/dt = d/dt(1/(t+1))
= -1/(t+1)^2
Now, we can apply the chain rule to find dy/dx:
dy/dx = (dy/dt)/(dx/dt)
= (-1/t^2) / (-1/(t+1)^2)
= (t+1)^2 / t^2
= (t^2 + 2t + 1) / t^2
So, dy/dx = (t^2 + 2t + 1) / t^2.
Is it
Find dy/dx for x=1/t+1 and y=1/t-1
or is it
Find dy/dx for x=1/ (t+1) and y=1/ (t-1)
Apologies for the confusion. The equations are:
x = 1/(t + 1) and y = 1/(t - 1)
To find dy/dx, we will take the derivatives of y and x with respect to t, and then apply the chain rule.
First, let's differentiate y with respect to t:
dy/dt = d/dt(1/(t - 1))
dy/dt = -1/(t - 1)^2
Now, let's differentiate x with respect to t:
dx/dt = d/dt(1/(t + 1))
dx/dt = -1/(t + 1)^2
Next, we can use the chain rule to find dy/dx:
dy/dx = (dy/dt)/(dx/dt)
dy/dx = (-1/(t - 1)^2)/(-1/(t + 1)^2)
dy/dx = (t + 1)^2/(t - 1)^2
Therefore, dy/dx = (t + 1)^2/(t - 1)^2.
To find dy/dx, we can use the chain rule. The chain rule states that if we have a composition of two functions, such as y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = dy/dg * dg/dx.
In this case, let's find dy/dx step by step.
Step 1: Find dg/dx
Given x = 1/(t+1), we need to find the derivative of x with respect to t. To do this, we can rewrite x as x = (t+1)^(-1). Now, we can differentiate x with respect to t using the power rule:
dg/dx = -1 * (t+1)^(-1-1) = -1/(t+1)^2
Step 2: Find dy/dg
Given y = 1/(t-1), we need to find the derivative of y with respect to g. Since g = 1/(t+1), we can rewrite y = 1/g. Now, we differentiate y with respect to g:
dy/dg = -1/g^2 = -1/(1/(t+1))^2 = -(t+1)^2
Step 3: Apply the chain rule
Using the chain rule, we multiply dy/dg by dg/dx to get the final expression for dy/dx:
dy/dx = dy/dg * dg/dx = -(t+1)^2 * -1/(t+1)^2 = (t+1)^2/(t+1)^2 = 1
Therefore, dy/dx = 1.
To find dy/dx, we need to differentiate both x and y with respect to t and then use the chain rule to find \(\frac{{dy}}{{dx}}\).
Given:
\(x = \frac{1}{{t + 1}}\)
\(y = \frac{1}{{t - 1}}\)
Differentiate both equations with respect to t:
Differentiating \(x\) with respect to \(t\):
\(\frac{{dx}}{{dt}} = \frac{{d}}{{dt}} \left(\frac{1}{{t + 1}}\right)\)
Differentiating \(y\) with respect to \(t\):
\(\frac{{dy}}{{dt}} = \frac{{d}}{{dt}} \left(\frac{1}{{t - 1}}\right)\)
To find \(\frac{{dx}}{{dt}}\) and \(\frac{{dy}}{{dt}}\), we can use the power rule for differentiating \(\frac{1}{u}\), where \(u\) is a function of \(t\):
\(\frac{{d}}{{dt}} \left(\frac{1}{u}\right) = -\frac{{1}}{{u^2}} \cdot \frac{{du}}{{dt}}\)
Applying this rule, we have:
\(\frac{{dx}}{{dt}} = -\frac{{1}}{{(t + 1)^2}} \cdot \frac{{d}}{{dt}}(t + 1) = -\frac{{1}}{{(t + 1)^2}}\)
\(\frac{{dy}}{{dt}} = -\frac{{1}}{{(t - 1)^2}} \cdot \frac{{d}}{{dt}}(t - 1) = -\frac{{1}}{{(t - 1)^2}}\)
Now, we have the derivative of both x and y with respect to t. To find \(\frac{{dy}}{{dx}}\), we will use the chain rule:
\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)
Substituting the derivatives we found earlier:
\(\frac{{dy}}{{dx}} = \frac{{-\frac{{1}}{{(t - 1)^2}}}}{{-\frac{{1}}{{(t + 1)^2}}}} = \frac{{t + 1)^2}}{{(t - 1)^2}}\)
Finally, substitute the given value of \(t\) (in this case, \(t = 1\)) to find the value of \(\frac{{dy}}{{dx}}\):
\(\frac{{dy}}{{dx}} = \frac{{(1 + 1)^2}}{{(1 - 1)^2}} = \frac{{4}}{{0}}\)
Since the denominator is zero, the value of \(\frac{{dy}}{{dx}}\) is undefined.