Q: What is the derivative of 2log 1 plus x?

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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.

The derivative of y = 1/3 x3 - 3x2 + 8x + 1/3 is x2 - 6x + 8. You can determine this for yourself by the rules. The derivative of a constant (e.g. 1/3) is 0. The derivative of xn for positive n (actually all nonzero n) is nxn-1. And if the derivative of f(x) is f'(x), then the derivative of k f(x) is k f'(x). Put all these together and you get the above result.

x^0 = 1 for all x. The derivative of 1 is always zero.

Derivative of natural log x = 1/x

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the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x

d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)

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))

(-x+tanx)'=-1+(1/cos2x)

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negative sin(x)

There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).

The derivative is 1/(1 + cosx)

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Derivative of 1/x 1/x = x-1 Take the derivative (-1)x(-1-1) = -x-2 = 1/x2

The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x

The anti-derivative of sqrt(x) : sqrt(x)=x^(1/2) The anti-derivative is x^(1/2+1) /(1/2+1) = (2/3) x^(3/2) The anti-derivative is 4e^x is 4 e^x ( I hope you meant e to the power x) The anti-derivative of -sin(x) is cos(x) Adding, the anti-derivative is (2/3) x^(3/2) + 4 e^x + cos(x) + C