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d/dt cot (t) dt = - cosec2(t)
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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 the derivative of t?
If it is with respect to t: 1 If it is with respect to some other variable (x for example): (dt)/(dx), which is literally read "the derivative of t with respect to x"
What is the derivative of cot x?
The derivative of cot(x) is -csc2(x).(Which is the same as -1/sin2(x).)
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
What expression is equivalent to cot t sec t?
Which expression is equivalent to cot t sec t
What is the derivative of x equals cot2 divided by t?
Assuming you want dx/dt and that the equation is x = cot(2) / t (i.e. cot(2) is a constant) we can use the power rule. First, we rewrite it: cot(2)/t = cot(2) * t-1 thus, by the power rule: dx/dt = (-1) cot(2) * t-1 -1 = - cot(2) * t-2= = -cot(2)/t2
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.
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.
What is the derivative of t?
If it is with respect to t: 1 If it is with respect to some other variable (x for example): (dt)/(dx), which is literally read "the derivative of t with respect to x"
What is the derivative of te(-t2)?
Use the product and chain rules: d/dt te^(-t²) = (d/dt t)e^(-t²) + t(d/dt e^(-t²)) = 1 × e^(-t²) + t × e^(-t²) × d/dt -t² = e^(-t²) (1 + t × -2t) = (1 - 2t²)e^(-t²)
What is the derivative of cot x?
The derivative of cot(x) is -csc2(x).(Which is the same as -1/sin2(x).)
What is the 4th derivative of the position function?
Distance f(x) = x(t) = 1/2at2+vit+xi Velocity f'(x) = v(t) = dx/dt=at +vi Acceleration f''(x) = a(t) = dv/dt Jerk f'''(x) = j(t) = da/dt The fourth derivative is not universally accepted, but some have called it "snap", and the fifth and sixth derivatives "crackle" and "pop" respectively. Snap f(4)(x) = s(t) = dj/dt Crackle f(5)(x) = c(t) = ds/dt Pop f(6)(x) = p(t) = dc/dt
What is the derivative of cot?
d/dx (cot x) = -csc2x
What is intantaeous speed?
If a body is moving at a non-constant speed then its instantaneous speed is the derivative of its displacement with respect to time.If the body is at positions x(t) and x(t+dt) at times t and t+dt then the instantaneous speed at time t is the limit, as dt tends to 0, of [x(t+dt) - x(t)]/t.In graph terms, it is the gradient of tangent to the displacement-time graph.
What is the derivative of cotx?
The derivative of cot(x) is -csc2(x).
How do you prove the derivative of parametric equations?
The question is to PROVE that dy/dx = (dy/dt)/(dx/dt). This follows from the chain rule (without getting into any heavy formalism). We know x and y are functions of t. Given an appropriate curve (we can integrate piece-wise if necessary), y can be written as a function of x where x is a function of t, i.e., y = y(x(t)). By the chain rule, we have dy/dt = dy/dx * dx/dt. For points where the derivative of x with respect to t does not vanish, we therefore have (dy/dt)/(dx/dt) = dy/dx.
What is acceleration of a particle if position of a particle at any instant of time t is given by x equals t 3?
velocity is 1st derivative of distance with respect to time acceleration is 2nd derivative of distance with respect to time dx/dt = velocity = 3t^2 dv/dt = acceleration = 6t
