The derivative is 1/(1 + cosx)
The derivative of x divided by 3 is 1/3. This can be found using the power rule of differentiation, where the derivative of x^n is nx^(n-1). In this case, x can be written as x^1, so the derivative is 1(1/3)*x^(1-1) = 1/3.
f(x)=xln(x) this function is treated as u*v u=x v=ln(x) The derivative of a product is f'(x)=u*v'+v*u' plugging the values back in you get: f'(x)=(x*dlnx/x)+(ln*dx/dx) The derivative of lnx=1/x x=u dlnu/dx=(1/u)*(du/dx) dx/dx=1 x=u dun/dx=nun-1 dx1/dx=1*x1-1 = x0=1 f'(x)=x*(1/x)+lnx*1 f'(x)=1+lnx Now for the second derivative f''(x)=d1/dx+dlnx/dx the derivative of a constant, such as 1, is 0 and knowing that the derivative of lnx=1/x you get f''(x)=(1/x)
(1/2(x^-1/2))/x
The derivative of 2/x can be found using the quotient rule in calculus. The quotient rule states that the derivative of f(x)/g(x) is [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2. Applying this rule to 2/x, where f(x) = 2 and g(x) = x, the derivative is calculated as [x0 - 21] / x^2, which simplifies to -2/x^2. Therefore, the derivative of 2/x is -2/x^2.
It is -1 over x-squared.
1/x = x-1d/dx(x-1) = -x-2 = -1/x2
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
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.
Derivative of x = 1, and since sqrt(x) = x^(1/2), derivative of x^(1/2) = (1/2)*(x^(-1/2))Add these two terms together and derivative = 1 + 1/(2*sqrt(x))
The anti-derivative of 1/x is ln|x| + C, where ln refers to logarithm of x to the base e and |x| refers to the absolute value of x, and C is a constant.
the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x
d(√(x)/5 ,x) = 1/( 10√(x))
The derivative is 1/(1 + cosx)
d/dx(1/2x) = -1/(2x2)
4/x can be written as 4x-1 (the power of negative 1 means it is the denominator of the fraction) 4*-1 = -4 Therefore, the derivative is -4x-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