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.
2.5x2 + any constant
9
f(x) = 3x2 + 5x + 2fprime(x) = 6x + 5
y = 5xy' = 5xln5 = ln5(5x)y" = ln5(5xln5) = (ln5)25x = 2.59(5x)
The derivative of 5x is 5.
x = 10x, so derivative = 10
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
Show the terms as additive here..., 8X2 + 10X3 factor out 2X2 2X2(4 + 5X) ---------------- 2X2 ==== greatest common factor
The idea is to use the addition/subtraction property. In other words, take the derivative of 5x, take the derivative of 1, and subtract the results.
The "double prime", or second derivative of y = 5x, equals zero. The first derivative is 5, a constant. Since the derivative of any constant is zero, the derivative of 5 is zero.
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.
10 x
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).
2.5x2 + any constant
9
20x -5