y = 5x
y' = 5xln5 = ln5(5x)
y" = ln5(5xln5) = (ln5)25x = 2.59(5x)
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
f(x)=(pi2)x=pi2x. The derivative of kx=ln(k)*kx, so f'(x)=2ln(pi)*pi2x (with chain rule).
I'm not sure what you're asking. The derivative of 6x is 6. The derivative of x^6 is 6x^5.
The idea is to use the chain rule. Look up the derivative of sec x, and just replace "x" with "5x". Then multiply that with the derivative of 5x.
If x is a function of time, t, then the second derivative of x, with respect to t, is the acceleration in the x direction.
The first derivative of e to the x power is e to the power of x.
Write sec x as a function of sines and cosines (in this case, sec x = 1 / cos x). Then use the division formula to take the first derivative. Take the derivative of the first derivative to get the second derivative. Reminder: the derivative of sin x is cos x; the derivative of cos x is - sin x.
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
2x is the first derivative of x2.
2x is the first derivative of x2.
d/dx(X^4) = 4X^3 ( first derivative ) d/dx(4X^3) = 12X^2 ( second derivative )
The function given is (f(x) = -x^2). The second derivative of a function, denoted as (f’'(x)), measures the concavity of the function. For the function (f(x) = -x^2), the first derivative (f’(x)) is (-2x). Taking the derivative of (f’(x)) gives us the second derivative (f’‘(x)), which is (-2). So, (f’'(x) = -2). This indicates that the function (f(x) = -x^2) is concave down for all (x), because the second derivative is negative.
2x
The fourth derivative of ( \ln(x) ) can be determined by first calculating its derivatives. The first derivative is ( \frac{1}{x} ), the second derivative is ( -\frac{1}{x^2} ), the third derivative is ( \frac{2}{x^3} ), and the fourth derivative is ( -\frac{6}{x^4} ). Thus, the fourth derivative of ( \ln(x) ) is ( -\frac{6}{x^4} ).
f(x)=(pi2)x=pi2x. The derivative of kx=ln(k)*kx, so f'(x)=2ln(pi)*pi2x (with chain rule).
Well if you have 5/X then you can rewrite this like 5x-1. And the derivative to that is -5x-2 and that can be rewrote to: -(5/x2).
y=x^pid/dx=pi*(x^pi-1)This is true because of power rule.d/dx (x^a)=a(x^(a-1))