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f(x)=|x|

|x| = sqrt(x^2)

f(x) = sqrt (x^2)

f'(x)=1/2sqrt(x^2) d/dx(x^2)

f'(x) = 2x /2sqrt(x^2)

f'(x) = x/sqrt(x^2)

f'(x)= x /|x|

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Note: d/dx [sqrt(x^2] = d/dx [x^(1/2)] =(1/2)x^(-1/2)=1/2x^(1/2)=1/2sqrt(x)

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Q: What is the first derivative of modulus function?
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What is the second derivative of a function's indefinite integral?

well, the second derivative is the derivative of the first derivative. so, the 2nd derivative of a function's indefinite integral is the derivative of the derivative of the function's indefinite integral. the derivative of a function's indefinite integral is the function, so the 2nd derivative of a function's indefinite integral is the derivative of the function.


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When the first derivative of the function is equal to zero and the second derivative is positive.


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If the first derivative of a function is greater than 0 on an interval, then the function is increasing on that interval. If the first derivative of a function is less than 0 on an interval, then the function is decreasing on that interval. If the second derivative of a function is greater than 0 on an interval, then the function is concave up on that interval. If the second derivative of a function is less than 0 on an interval, then the function is concave down on that interval.


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