The derivative of a constant is always 0. To show this, let's apply the definition of derivative.
Recall that the definition of derivative is:
f'(x) = lim h→0 (f(x + h) - f(x))/h
Let f(x) = 1. Then:
f'(x) = lim h→0 (1 - 1)/h
= lim h→0 0/h
= lim h→0 0
= 0!
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2.67676364564. Multiply the power of x by the coefficient, and lower the power of x by 1, which is x to the 0 power, or 1.
The derivative of ln(x) is 1/x. Therefore, by Chain Rule, we get:[ln(10x)]' = 1/10x * 10 = 1/xUsing this method, you can also infer that the derivative of ln(Ax) where A is any constant equals 1/x.
2.71828183 ==So the derivative of a constant is zero.If you have e^x, the derivative is e^x.
hi peoplehey therethe derivitiive of f at a is simply, the limit as h appraoches 0 of (f(a+h)-f(a))/h if this limit exists. QEDrockem
∫1/(1-x)2dxWe can rewrite the integral as ∫(1-x)-2dx.Thus:∫(1-x)-2dxu = 1-xdu = -1-1∫-1(1-x)-2dx-1∫u-2du(u-1|0u=1-x(1-x)-1So we conclude that ∫1/(1-x)2dx = (1-x)-1 or 1/(1-x)