In general, if you're taking the derivative with respect to X, then you take the current power of X, multiply the given quantity by that number and then subtract one from the current power. In this case, that's an overcomplicated way of describing what happens but here's the process: 5x is more fully 5*x^1 So you take the power (1) and multiply it it by the given quantity. This gives you 1*5*x^1 Now you subtract one from the current power giving you 1*5*x^0 which equals 5. So the answer is simply 5 in this case. But what if you were trying to find the derivative of 4x^7? In this case, you would multiply the quantity by 7 (giving you 7*4*x^7) and subtract 1 from the current power giving you a final answer of 28*x^6. This also works for negative powers and square roots. The derivative of sqrt(x) can be found by recognizing that this is equal to x^(1/2). So you multiply everything by 1/2 and subtract one from the power and get 1/2 * x^(-1/2) which equals 1/2 * 1/sqrt(x) = 1/(2*sqrt(x))
I understand this expression to be 10x3 + 5x. Derivative: 30x2 + 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.
2.5x2 + any constant
9
f(x) = 3x2 + 5x + 2fprime(x) = 6x + 5
The derivative of 5x is 5.
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).
2.5x2 + any constant