What is the derivative of Ae^kt*cos(wt+b)?
What is the derivative of x^(x^2+1)?
use the product rule. d/dt uv = u'v + uv'
d/dt A e^(kt) cos(wt+b)
= A*k e^(kt) cos(wt+b) + A*e^(kt)(-w sin(wt+b))
= Ae^(kt) [k cos(wt+b) - w sin(wt+b)]
For the other, start by noting that if
y = u^v
ln y = v ln u
1/y y' = v' lnu + v * 1/u u'
y' = y/u u' + ln u y v') = v u^(v-1) u' + ln u u^v v'
This is just a combination of the power rule (d/dx u^n = n u^(n-1) u')
and the exponent rule (d/dx a^v = lna a^v v')
whew ... so
d/dx x^(x^2+1) = (x^2+1) x^(x^2) + 2x lnx x^(x^2+1)
or, if you like, x^(x^2+1) (x + 1/x + 2x lnx)
To find the derivative of Ae^kt*cos(wt + b), we will need to apply the product rule and the chain rule.
Let's start with the product rule. The product rule states that if we have two functions being multiplied together, the derivative of the product is given by:
(d/dx)(f(x) * g(x)) = f'(x) * g(x) + f(x) * g'(x)
In this case, our two functions are Ae^kt and cos(wt + b):
Let's find the derivative of the first function Ae^kt:
(d/dt)(Ae^kt) = A * (d/dt)(e^kt)
Now, let's find the derivative of e^kt. The derivative of e^kt with respect to t is simply k * e^kt:
(d/dt)(e^kt) = k * e^kt
So, the derivative of Ae^kt is A * k * e^kt.
Now, let's move on to finding the derivative of the second function cos(wt + b):
(d/dt)(cos(wt + b))
The derivative of cos(wt + b) with respect to t can be found using the chain rule. The chain rule states that if we have a composite function, the derivative is given by the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
In this case, the outer function is cos(u), where u = wt + b. The inner function is wt + b.
The derivative of the outer function cos(u) with respect to u is -sin(u), and the derivative of the inner function wt + b with respect to t is w.
Therefore, the derivative of cos(wt + b) is -sin(wt + b) * w.
Now, applying the product rule, we can find the derivative of the entire function Ae^kt*cos(wt + b):
(d/dt)(Ae^kt*cos(wt + b))
= (A * k * e^kt) * cos(wt + b) + Ae^kt * (-sin(wt + b) * w)
= A * k * e^kt * cos(wt + b) - A * w * e^kt * sin(wt + b)
So, the derivative of Ae^kt*cos(wt + b) is A * k * e^kt * cos(wt + b) - A * w * e^kt * sin(wt + b).
Now, let's move on to finding the derivative of x^(x^2+1).
To find the derivative of x^(x^2+1), we can use logarithmic differentiation.
Let y = x^(x^2+1).
Take the natural logarithm of both sides: ln(y) = ln(x^(x^2+1)).
Using the power rule of logarithms, we can simplify the right-hand side: ln(y) = (x^2+1) * ln(x).
Now, differentiate both sides with respect to x:
(d/dx)(ln(y)) = (d/dx)((x^2+1) * ln(x)).
Using the chain rule, the left-hand side simplifies to: (1/y) * (dy/dx).
Using the product rule, the right-hand side simplifies to: (2x * ln(x) + (x^2+1) * (1/x)).
Now, multiply both sides by y:
(1/y) * (dy/dx) = 2x * ln(x) + (x^2+1) * (1/x).
Substitute y = x^(x^2+1) back in:
(1/x^(x^2+1)) * (dy/dx) = 2x * ln(x) + (x^2+1) * (1/x).
Multiply both sides by x^(x^2+1):
dy/dx = (x^(x^2+1)) * (2x * ln(x) + (x^2+1) * (1/x)).
So, the derivative of x^(x^2+1) is (x^(x^2+1)) * (2x * ln(x) + (x^2+1) * (1/x)).