Zero. In general, the derivative of any constant is zero.
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Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
"Derivative of"
8.3
I am interpreting this as: Find the derivative of: y=((2x+5)8+4x)3 To find the derivative (y'), the chain rule must be applied. The "outermost" function of this compound function is t3 (t being an arbitrary quantity). The derivative of t3 is 3t2 * dt, where "dt" is the derivative of the quantity "t". Applying this, we arrive at a working definition of y': y' = 3((2x+5)8+4x)2(derivative of (2x+5)8+4x) The derivative of (2x+5)8+4x is found using basic derivative definitions and the chain rule again: 8(2x+5)7(2)+4 = 16(2x+5)7+4 So now we can write y' again: y'= 3((2x+5)8+4x)2(16(2x+5)7+4) = 48((2x+5)8+4x)2((2x+5)7+4) This can be further simplified, but this is an arduous process. If you need further simplification, feel free to contact me via private message.
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.