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If derivative of f equals 0 for all x in the domain of f then is f a constant function?
Yes.
How do you find the derivative of f of x equals 6 over x plus 3?
The original equation f(x) = 6/(x+3) can be rewritten as f(x) = 6(x+3)-1. Now derive the equation according the the power rule and the chain rule: y = 6 (x+3)-1 dy/dx = 6 (-1)(x+3)-2(1)* dy/dx = -6/(x+3)2 * by the chain rule, you must multiply by the derivative of which is simply one. Thus, the derivative of f(x) = 6/(x+3) equals -6/(x+3)2
What is the derivative of F of x equals 7x squared plus 5?
14x
What is the derivative of F of x equals x plus 1 divided by x?
x= (x+1)/x(x+1)/x -x = 0f'(x) = ((x*1)-(x+1)) /x^2 -11/x^2 - 1
What is the first derivative of a mod b?
The derivative of f(x) = x mod b is f'(x)=1, except where x is a multiple of b, when it is undefined.
What is the derivative for x0?
f'(x)= 0x^-1=0anything multiplied by zero equals zero
If derivative of f equals 0 for all x in the domain of f then is f a constant function?
Yes.
How do you find the derivative of f of x equals 6 over x plus 3?
The original equation f(x) = 6/(x+3) can be rewritten as f(x) = 6(x+3)-1. Now derive the equation according the the power rule and the chain rule: y = 6 (x+3)-1 dy/dx = 6 (-1)(x+3)-2(1)* dy/dx = -6/(x+3)2 * by the chain rule, you must multiply by the derivative of which is simply one. Thus, the derivative of f(x) = 6/(x+3) equals -6/(x+3)2
What is the derivative of ln 1-x?
In this case, you need to apply the chain rule. Note that the derivative of ln N = 1/N. In that case we get: f(x) = ln(1 - x) ∴ f'(x) = 1/(1 - x) × -1 ∴ f'(x) = -1/(1 - x)
What is the derivative of F of x equals 7x squared plus 5?
14x
What is the derivative of modulus of x?
Well.. this is the formula to get the derivative of the modulus - d|f(x)|/dx = [ |f(x)|/f(x) ] * f'(x)
What is the second derivative of x ln x?
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)
What is the derivative of F of x equals x plus 1 divided by x?
x= (x+1)/x(x+1)/x -x = 0f'(x) = ((x*1)-(x+1)) /x^2 -11/x^2 - 1
What is the first derivative of a mod b?
The derivative of f(x) = x mod b is f'(x)=1, except where x is a multiple of b, when it is undefined.
What is the derivative of arccosxdividedby2?
f(x) = arccos(x) / 2 f'(x) = -1/(2√(1 - x2))
What is the derivative of 1 divided by fx under the conditions that fx is differentiable and fx cannot be 0?
Negative the derivative of f(x), divided by f(x) squared. -f'(x) / f²(x)
Can you please help me with this hard higher maths question on differentiation find the derivative of f of x equals xdivided by 4 plus 4divided by x?
f(x)=x/4 + 4/x f'(x) = 1/4 - 4/x^2
