hollabackguru Posted October 16, 2018 Report Share Posted October 16, 2018 1. Prove that 1/(2x+1) = -2/(2x+1)^2 using first principles. 2. Given the equation x^2 - 5, find the derivative of the function using first principles and then find the tangent. Reply Link to post Share on other sites More sharing options...
literaturegirl Posted January 19, 2019 Report Share Posted January 19, 2019 On 10/16/2018 at 8:51 AM, hollabackguru said: 1. Prove that 1/(2x+1) = -2/(2x+1)^2 using first principles. 2. Given the equation x^2 - 5, find the derivative of the function using first principles and then find the tangent. 1. Use first principles formula. However, quotient rule is quicker alternative... 2. First you find the slope function of f(x) = x^2 - 5, which is parabola. To find the tanget, you need to know the point where they're asking you for the tangent, any point as (x,y) for example (1,2). As the slope of function f(x) x^2 - 5 is f´(x) = 2x, just plug in the x-value of the point (which is 1 in this case) --> 2 x 1 = 2. That's the SLOPE of the parabola at point (1,2). Hence, you can use the line equation: y - y0 = m (x - x0), where m = slope at the point, so 2. Substitute point (1,2) to function as x0 = 1 and y0 = 2 ------> y - 2 = 2(x-1) and calculate it to get nice equation. 1 Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 19, 2019 Report Share Posted January 19, 2019 @literaturegirl Are you sure that (1, 2) is on the graph of f(x) = x^2 - 5? Reply Link to post Share on other sites More sharing options...
literaturegirl Posted January 19, 2019 Report Share Posted January 19, 2019 44 minutes ago, kw0573 said: @literaturegirl Are you sure that (1, 2) is on the graph of f(x) = x^2 - 5? It’s not. Used some random point as an example as the original question didn’t mention any. Reply Link to post Share on other sites More sharing options...
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