Hello!!
1. find the third derivative of
f(t)=(t-1)/(t+1)
2.
y= x^(x^2) ; dy/dx
please help! thank you!
f = (t-1)/(t+1)
f' = 2/(t+1)^2
now it's easy
f'' = -4/(t+1)^3
f''' = 12/(t+1)^4
y = x^(x^2)
lny = x^2 ln x
1/y y' = 2x lnx + x^2/x
y' = y(2x ln x + x)
and plug back in y = x^(x^2)
Hello! I'd be happy to help you with your questions.
1. To find the third derivative of f(t) = (t-1)/(t+1), we need to apply the chain rule multiple times. Here are the steps to find the third derivative:
First, let's find the first derivative:
f'(t) = [(t+1)(1) - (t-1)(1)] / (t+1)^2
= (2) / (t+1)^2
Next, let's find the second derivative:
f''(t) = [0 - (2)(2)(t+1)] / (t+1)^4
= (-4(t+1)) / (t+1)^4
= -4 / (t+1)^3
Finally, let's find the third derivative:
f'''(t) = [(0)(t+1)^3 - (-4)(3)(t+1)^2] / (t+1)^6
= (12(t+1)^2) / (t+1)^6
= 12 / (t+1)^4
So, the third derivative of f(t) = (t-1)/(t+1) is f'''(t) = 12 / (t+1)^4.
2. To find dy/dx for y = x^(x^2), we need to use logarithmic differentiation. Here are the steps to find the derivative:
Let's take the natural logarithm of both sides:
ln(y) = ln(x^(x^2))
Using a logarithmic property, we can rewrite the equation as:
ln(y) = (x^2) ln(x)
Now, let's differentiate both sides with respect to x:
(d/dx) ln(y) = (d/dx) [(x^2) ln(x)]
Using the chain rule on the right side, we have:
[1/y] (dy/dx) = 2x ln(x) + x^2 * (1/x)
= 2x ln(x) + x
Finally, let's solve for dy/dx by multiplying both sides by y:
dy/dx = y * [2x ln(x) + x]
To find y, we can substitute the original equation y = x^(x^2):
dy/dx = x^(x^2) * [2x ln(x) + x]
So, the derivative of y = x^(x^2) is dy/dx = x^(x^2) * [2x ln(x) + x].
I hope that helps! If you have any more questions, feel free to ask.