given, y = x^2+2, u = 2*X-1, THEN dy/du = can someone help me with this problem, thanks
G Guest Guest Dec 20, 2005 #1 given, y = x^2+2, u = 2*X-1, THEN dy/du = can someone help me with this problem, thanks
H happy Full Member Joined Oct 30, 2004 Messages 466 Dec 20, 2005 #2 given, y = x^2+2, u = 2*X-1, THEN dy/du = Is it (x^2) +2 or x^(2+2). Is it (2*x)-1 or 2*(x-1)?
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Dec 20, 2005 #3 G'day, Bluemega. Given, \(\displaystyle y = x^2+2\), \(\displaystyle u = 2x - 1\), then \(\displaystyle \L \frac{dy}{du} = ?\) Click to expand... We can use the chain rule: . . .\(\displaystyle \L \frac{dy}{du} = \frac{dy}{dx} \times \frac{dx}{du}\) or better written here as . . . \(\displaystyle \L \frac{dy}{du} = \frac{dy}{dx} / \, \frac{du}{dx}\) [1] We have . . .\(\displaystyle \L y = x^2 + 2\) . . . . so \(\displaystyle \L \frac{dy}{dx} = 2x\) Find \(\displaystyle \L \frac{du}{dx}\) similarly, apply [1] and you have \(\displaystyle \L \frac{dy}{du}\).
G'day, Bluemega. Given, \(\displaystyle y = x^2+2\), \(\displaystyle u = 2x - 1\), then \(\displaystyle \L \frac{dy}{du} = ?\) Click to expand... We can use the chain rule: . . .\(\displaystyle \L \frac{dy}{du} = \frac{dy}{dx} \times \frac{dx}{du}\) or better written here as . . . \(\displaystyle \L \frac{dy}{du} = \frac{dy}{dx} / \, \frac{du}{dx}\) [1] We have . . .\(\displaystyle \L y = x^2 + 2\) . . . . so \(\displaystyle \L \frac{dy}{dx} = 2x\) Find \(\displaystyle \L \frac{du}{dx}\) similarly, apply [1] and you have \(\displaystyle \L \frac{dy}{du}\).
