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Can you use the equation given below to find the second derivative of pi divided by 6 if fx equals cscx?
pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!
How do you get the second derivative of g of x equals xcscx where x equals theta?
T=theta so that it will not look so messy. g(T)=TcscT To find the first derivative, you must use the product rule. Product rule is derivative of the first times the second, plus the first times the derivative of the second, which will give you: g'(T)=0xcscT + Tx-cscTcotT, which simplifies: g'(T)= -cscTxcotT Now, take the derivative of that to get the second derivatice. In order to do that, you have to do the product rule again. g"(T)=(cscTcotT)cotT + -cscT(-csc^2T) {that's csc squared} which simplifies: g"(T)= cscTcot^2(T) + csc^3 (T)
Find the derivative y equals sq rt 7 plus 99 plus pie?
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.
How do you calculate critical points of derivatives?
The "critical points" of a function are the points at which the derivative equals zero or the derivative is undefined. To find the critical points, you first find the derivative of the function. You then set that derivative equal to zero. Any values at which the derivative equals zero are "critical points". You then determine if the derivative is ever undefined at a point (for example, because the denominator of a fraction is equal to zero at that point). Any such points are also called "critical points". In essence, the critical points are the relative minima or maxima of a function (where the graph of the function reverses direction) and can be easily determined by visually examining the graph.
Formulae in derivative?
You will find several formulae in the Wikipedia article on "derivative".
