Help with derivative!?!?
wow ok this is pretty straight forward. f(x) = x^3 f '(x) = 3x^2 f '(2) = 3(2)^2 f '(2) = 12 f(x) = sqrt of x rewrite it as x^1/2 so the derivative is... f '(x) 1/2x^3/2 lastly the derivative of the last one is f(x) = x^2 + 3x + 2 f '(x) = 2x + 3 f '(2) = 7 you put the value of 2 into the derivative for the ones asking for f '(2)
Need help with derivative?
f(x) = (- 5x + 6) / (x + 1) = [ (- 5x - 5) + 11 ] / (x + 1) = [ (- 5(x + 1) + 11 ] / (x + 1) = - 5 + 11 / (x + 1). f'(x) = - 11 / (x + 1)^2.
Help with derivatives?
f(5) = 700 -15 = 685 f'(t) = -3 <=> f'(5) = -3 => annual decreasing Example : f(5) = 685 f (4) = 700 - 12 = 688 688 - 685 = 3 (slope of the line = annual decreasing)
Help with derivatives of unit vectors?
x_hat,y_hat,and z_hat, are what is called, CARTESIAN COORDINATES. These make up a coordinate system, that doesn't change with the situation of the problem, with time. They all point in fixed directions. r_hat and theta_hat, are polar coordinates. r_hat points from a center point, out towards the point you are interested in. theta_hat, points around a circle, that you'd circumnavigate from a standard direction, to get to your point of interest. I'm not exactly sure what you mean by "omega_hat". Traditionally, omega refers to the rate of change in theta, also known as angular velocity. To get derivatives of changing unit vectors, it helps to express them in terms of the non-changing unit vectors. Then do your differentiation, so you can treat the non-changing unit vectors (Cartesian unit vectors) as constants. for r_hat: r_hat = cos(theta)*x_hat + sin(theta)*y_hat theta being an angle around the circle, to which your radial direction points. The theta_hat direction continues that direction of tracing the angle around the circle. Thus, it is perpendicular to r_hat. Draw out the picture of it, and you will see that you negate the sine term, swap it to the x, and then swap the cosine term to the y. theta_hat = -sin(theta)*x_hat + cos(theta)*y_hat
Calculus Help w/ derivative?
f ' (x) = (-4x/5)sin(x^2/10) 1. dy/dx = 2sec(x)tan(x) + 3sec^2(x) 2. dy/dx = 4 3. dy/dx = 8tan^3(x)sec^2(x)
What are the reasons for the use of derivatives?
What is Derivative Market?Derivative is a financial contract whose value is derived from underlying assets which can be stocks, commodities, bonds, currencies, market indexes or interest rates. Derivatives can be either traded through exchange or Over the Counter (OTC).Read: A Detailed Understanding of Derivative MarketDerivatives help in making profit by simply betting on the future value of the underlying assets.Derivatives helps investors in:Hedging your securitiesIt can be used to hedge our securities from fluctuating of market prices. The shares which we posses can be protected on the downside by entering into derivative contractTransferring of risksIt helps in transferring of risk from risk averse people to risk seekers. The risk averse traders can secure their positions by entering into derivative contract. Whereas risk seekers can enter into risky trades and make short term profitsBenefiting from arbitrage opportunitiesArbitrage means taking advantage of the price difference in both the markets.One can buy at a lower from one market and sell at a higher price in another market.With the help of the derivative contract, one can take advantage of the price different in the two marketYou can also learn more about derivatives by watching the video below:
Calculus: For the love of derivatives- help me!?
I usually find these things look a bit less intimidating with the powers of things put in rather than use Sqrt and 1-over bars A) Y=Ln[(x+1) * (2x-7)^-1] U=(x+1) dv=-2(2x-7)^-2 V=(2x-7)^-1 du=1 Now do (U*dv)+(V*du) to get the interior derivative [(x+1) * (-2(2x-7)^-2)] + [(2x-7)^-1] Multiply that by the derivative of Log(Fn) which just means put the original function under a bar with that interior derivative mess on top: [(x+1) * (-2(2x-7)^-2)] + [(2x-7)^-1] --------------------------------------... (X+1)*((2x-7)^-1) This then breaks into two fractions: [(x+1) * (-2(2x-7)^-2)] ----------------------------] (X+1)*((2x-7)^-1) Plus [(2x-7)^-1] -------------] (X+1)*((2x-7)^-1) With a bit of cancellation you get: Answer: [(x+1)^-1]-[2(2x-7)^-1] B) y = [Exp(3x)-Log(x^4)]^3 dExp(3x)=3Exp(3x) dLog(X^4)=(4/x) Answer: 3([Exp(3x)-Log(x^4)]^2) * (3Exp(3x) - (4/x)) C) Y= 7( 7Log(x)^0.5)^-1 d7Log(x)^0.5 = (7/2)(Log(x)^0.5)(1/x) Gives: [7 / (2x (Log(x)^0.5)] * [-7/((7Log(x)^0.5)^2)] Reduces to: [7 / (2x Log(x)^0.5)] * [-7 / (49 Log(x)] Reduces to: -7 / [2x (Log(x)^0.5) * 7 Log(x)] Answer: -1 / [2x Log(x)^1.5]