0
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
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.
Can the concavity of a curve be determined by differentiation?
Yes, the concavity of a curve can be determined by differentiation. To find out the concavity of a graph at various points, you want to analyze the second derivative (f''(x)). Take the derivative of your original equation, then, take the derivative of this equation. By setting this second derivative to zero, you can solve for the critical points (x-intercepts/asymptotes) of the second derivative graph. Once these critical points are found, make a number line with these points marked. By doing a sign test on either sides of the critical points (plug in numbers below and above the critical points into the second derivative equation), you can find the concavities of your original graph. Wherever the sign tests results in a positive number, that is where a upward facing curve is (concave up); where it is negative, that is where a concave down portion is.
How do you find the mean of x over y equals a over b?
Find the derivative of Y and then divide that by the derivative of A
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 find second derivative of a function?
All it means to take the second derivative is to take the derivative of a function twice. For example, say you start with the function y=x2+2x The first derivative would be 2x+2 But when you take the derivative the first derivative you get the second derivative which would be 2
How do you find the second derivative?
Afetr you take the first derivative you take it again Example y = x^2 dy/dx = 2x ( first derivative) d2y/dx2 = 2 ( second derivative)
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)
Where the graph of a function equals the value zero?
you have to first find the derivative of the original function. You then make the derivative equal to zero and solve for x.
How do you find the derivative of y equals sin8x?
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
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 find the ponts of inflexion on a curve?
At the point of inflexion:the first derivative must be zero. the second derivative must be zero, if the next derivative is zero then the one following that must also be zero.
Find the derivative of y equals 3cosx?
y=3 cos(x) y' = -3 sin(x)
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.
How do you find the equation of a tangent line to the graph of fx equals x cubed and x equals -2?
Calculate the derivative of the function.Use the derivative to calculate the slope at the specified point.Calculate the y-coordinate for the point.Use the formula for a line that has a specified slope and passes through a specified point.
