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.
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.
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)
10 x