95x?
d/dx(au)=au*ln(a)*d/dx(u)
d/dx(95x)=95x*ln(9)*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(95x)=95x*ln(9)*(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(95x)=95x*ln(9)*(5*1)
d/dx(95x)=95x*ln(9)*(5)
-95x can simplify to (95)x, which equals 59049x.
-ln(9) can simplify to ln(32), so you can take out the exponent to have 2ln(3).
d/dx(95x)=59049x*2ln(3)*(5)
d/dx(95x)=10*59049x*ln(3)
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.
find anti derivative of f(x) 5x^4/3 + 8x^5/4
10 x
The derivative of f(x) is lim h-->0 [f(x+h)-f(x)]/h. So let f(x) = -5x. The derivative is lim h-->0 [-5(x+h)- -5(x)]/h = lim h-->0 [-5x - 5h + 5x]/h = lim h-->0 -5h/h Since the limit h-->0 of h/h is 1, the derivative is -5
The derivative of 5x is 5.
9
I understand this expression to be 10x3 + 5x. Derivative: 30x2 + 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)
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.
find anti derivative of f(x) 5x^4/3 + 8x^5/4
10 x
The derivative of f(x) is lim h-->0 [f(x+h)-f(x)]/h. So let f(x) = -5x. The derivative is lim h-->0 [-5(x+h)- -5(x)]/h = lim h-->0 [-5x - 5h + 5x]/h = lim h-->0 -5h/h Since the limit h-->0 of h/h is 1, the derivative 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).