I'm not sure if you are asking about: e-ax2 or e-ax^2. Use the carets=^ in the future to raise something to a power.
For e-ax2:
You need to remember that: (ex)`= ex
also: (eax)`= aeax, assuming a is a constant and not a function of x.
Justrecognizeany constant drops down to the front.
Therefore: (e-ax2)`=-2ae-ax2, treat the negative just as a constant or (-1)
For e-ax^2: this one is a little bit trickier:
We can just use u-substitution.
Let u = x2.
u` = 2x
If we substitute the u for x^2, we have: e-au , and we know the deriviate of that is -ae-2au, but we need to multiply by u` as well to get:
(-ae-2au)(2x) = -2xae-2au, then we can just plug u back in.
u = x2, so:
-2xae-2au = -2xae-2ax^2
And that is the answer.
(e-ax^2)`= -2xae-2ax^2
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dy/dx = 3 integral = (3x^2)/2
y = Sin(x) dy/dx = Cos(x)
log 3 is a constant, so d/dx log3, like d/dx of any constant, equals zero.
The open channel flow has a free surface whereas the pipe flow has a closed surface.
The derivative of the ln function is the function 1/x. So the derivative of ln1.01 should be 1/1.01 = 0.990099... ------------------------- Well I may be looking at this slightly different, but the question as stated "differentiate ln(1.01)" would be 0 seeing as ln(1.01) is itself a constant (irrational) number. The derivative of any constant is zero. If the intended question was ln(x)d/dx where x=1.01 then I agree with the above answer.