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What is derivative of 9et?
Assume that the expression is: y = 9e^(t) Remember that the derivative of e^(t) with respect to t is e^(t). If we take the derivative of the function y, we have.. dy/dt = 9 d[e^(t)]/dt = 9e^(t) Note that I factor out the constant 9. If we keep the 9 in the brackets, then the solution doesn't make a difference.
What is derivative of cot t dt?
d/dt cot (t) dt = - cosec2(t)
What is the value for acceleration in the x direction?
If x is a function of time, t, then the second derivative of x, with respect to t, is the acceleration in the x direction.
Prove that a constant vector always has a perpendicular derivative?
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
What is the derivative of e7x?
The derivative of e7x is e7 or 7e.The derivative of e7x is 7e7xThe derivative of e7x is e7xln(7)
Derivative of x to the power -t?
well if you're finding the derivative with respect to x, it would be -tx^(-t-1)
What is the derivative of t t-1?
-t/(t-1)^2+1/(t-1)
What is the Anti-derivative of 4et-sint?
4e^t+cos(t)
What is derivative of 9et?
Assume that the expression is: y = 9e^(t) Remember that the derivative of e^(t) with respect to t is e^(t). If we take the derivative of the function y, we have.. dy/dt = 9 d[e^(t)]/dt = 9e^(t) Note that I factor out the constant 9. If we keep the 9 in the brackets, then the solution doesn't make a difference.
What is derivative of cot t dt?
d/dt cot (t) dt = - cosec2(t)
Prove that if vector A has constant magnitude then its derivative is perpendicular to vector A?
If vector A has constant magnitude, then its derivative will be tangent to the direction of vector A. Since the derivative is perpendicular to the tangent vector, it will be perpendicular to vector A. This is because the derivative represents the rate of change of vector A with respect to time, which is perpendicular to the direction of a vector with constant magnitude.
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)
What is the Laplace transform of the third derivative of t squared?
Can you be more specific.
What is the instantaneous acceleration a of the particle at t45.0s Hints?
To find the instantaneous acceleration at t = 45.0s, you need to differentiate the velocity function with respect to time. The acceleration at t = 45.0s is the derivative of the velocity function at that time. Apply the derivative to the velocity function to find the acceleration at t = 45.0s.
What is the formula for instantaneous acceleration?
The formula for instantaneous acceleration is given by the derivative of velocity with respect to time: a(t) = dv(t) / dt, where a(t) is the acceleration at time t and v(t) is the velocity at time t.
What is the value for acceleration in the x direction?
If x is a function of time, t, then the second derivative of x, with respect to t, is the acceleration in the x direction.
What is the derivative of the square root of x to the seventh power?
7/2 t^5/2^
