Rprime= 2q + 2a + 3
Rdoubleprime= 4
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)
14x
first derivative = 6x2 + 6x - 36 factors are (6x + 18)(x - 2)
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0
The first derivative is m and the second is 0 so the third is also 0.
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)
x = 10x, so derivative = 10
2 and 3
14x
first derivative = 6x2 + 6x - 36 factors are (6x + 18)(x - 2)
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.
Following the correct order of operations: derivative of x^2 + 6/2 = derivative of x^2 +3, which equals 2x
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
if y=aex+be2x+ce3x then its derivative dy/dx or y' is: y'=aex+2be2x+3ce3x
Multiply the first by 8 and the second by 3