25x?
d/dx(au)=au*ln(a)*d/dx(u)
d/dx(25x)=25x*ln(2)*d/dx(5x)
-The derivative of 5x is:
d/dx(cu)=c*du/dx where c is a constant
d/dx(5x)=5*d/dx(x)
d/dx(25x)=95x*ln(2)*(5*d/dx(x))
-The derivative of x is:
d/dx(x)=1x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
d/dx(25x)=25x*ln(2)*(5*1)
d/dx(25x)=25x*ln(2)*(5)
-25x can simplify to (25)x, which equals 32x.
d/dx(95x)=32x*ln(2)*(5)
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
The derivative of 5x is 5.
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).
I understand this expression to be 10x3 + 5x. Derivative: 30x2 + 5.
9
x = 10x, so derivative = 10
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.
f(x) = 3x2 + 5x + 2fprime(x) = 6x + 5
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.
ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)
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.
2.5