When any number is multiplied by its reciprocal, the product is ' 1 '.
If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)
It increases the value.
The derivative of xe is e. The derivative of xe is exe-1.
Oh, dude, the third derivative of ln(x) is -2/(x^3). But like, who really needs to know that, right? I mean, unless you're planning on impressing your calculus teacher or something. Just remember, math is like a puzzle, except no one actually wants to put it together.
The derivative of ln x is 1/x. Replacing the expression, that gives you 1 / (1-x). By the chain rule, this must then be multiplied by the derivative of (1-x), which is -1. So, the final result is -1 / (1-x).
f'(x)= 0x^-1=0anything multiplied by zero equals zero
When any number is multiplied by its reciprocal, the product is ' 1 '.
If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)
Given y=ln(1/x) y'=(1/(1/x))(-x-2)=(1/(1/x))(1/x2)=x/x2=1/x Use the chain rule. The derivative of ln(x) is 1/x. Instead of just "x" inside the natural log function, it's "1/x". Since the inside of the function is not x, the derivative must be multiplied by the derivative of the inside of the function. So it's 1/(1/x) [the derivative of the outside function, natural log] times -x-2=1/x2 [the derivative of the inside of the function, 1/x] This all simplifies to 1/x So the derivative of ln(1/x) is 1/x
It is negative one divided by 4 multiplied by x to the power of 1.5 -1/(4(x^1.5))
The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.
the derivative of 1x would be 1 the derivative of x to the power of 1 would be 1. the derivative of x+1 would be 1 the derivative of x-1 would be 1 im not sure what you are asking, but however you put it, it's 1.
The rule in differentiating any bracket is: The power of the bracket multiplied by the bracket itself but lowered by one degree, then multiplied by the derivative of whatever inside the bracket. Its easy, just remember it this way. y=(1-x^3)^3 y'= 3(1-x^3)^2(-3x^2)= -9x^2(1-x^3)^2
It increases the value.
The derivative of xe is e. The derivative of xe is exe-1.
Derivative of 1/x 1/x = x-1 Take the derivative (-1)x(-1-1) = -x-2 = 1/x2