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What else can I help you with?
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)
Is the derivative of u with respect to v constant when t is held constant?
No, the derivative of u with respect to v is not constant when t is held constant.
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?
Suppose A is a vector with real components. A can be written as <f(t), g(t), h(t)>. Since the magnitude of A is constant we have f(t)*f(t) + g(t)*g(t) + h(t)*h(t) = c, where c is a non-negative real number. Take derivative of both sides of equation we get 2*f(t)*df(t)/dt + 2*g(t)*dg(t)/dt + 2*h(t)*dh(t)/dt = 0. Divide both sides by 2, we get f(t)*df(t)/dt + g(t)*dg(t)/dt + h(t)*dh(t)/dt = 0. Thus the dot product of A and its derivative is 0. This implies the angle between A and its derivative is Pi/2. Hence they are perpendicular.
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 derivative of the square root of x to the seventh power?
7/2 t^5/2^
