14)Find an equation of the tagent line to the graph of f(x)=x^2-2x-3 at the point (-2,5)
16)Find dy/dx for y=x(x+1)^1/3 -I rewrote x+1^1/3
17)Fin dy/dx if x^2+y^2=2xy
18)Differentiate y=sex^2x+tan^2x also do, y=2x/1-3x^2
take the derivative: 4x -2 (power rule) put -2 in for x, like given in problem. so... 4(-2)-2 = -10 that would be the slope of the line
equation of line
y - 5 = -10(x + 2) or y= -10x-15
y=sec^2 (x)+tan^2 (x) y'=2sec (x).[sec(x)]'+2tan (x).[tan(x)]' y'=2sec (x).[sec(x)tan(x)]+2tan (x).[sec^2(x)] y'=2sec^2 (x).tan(x)+2tan (x).sec^2(x) y'=4sec^2 (x).tan(x)
(Using Quotient Rule) which is d/dx [f(x) / g(x)] = [f '(x)g(x) - f(x)g'(x)] / g(x)^2 y'=[2(1 - 3x^2) - (2x)(-6x)] / (1-3x^2)^2 y'=[2 - 6x^2 + 12x^2] / 1-6x^2+ 9x^4 y'=[2+6x^2] / 9x^4-6x^2+1
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