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What is the derivative of 1 over X square plus 2 with respect to X using the first principle?
I'm not sure if you're asking what the derivative is for f(x) = 1/[(x^2)+2] or for f(x) = [1/(x^2)]+2 so I'm gonna do the first one for now. Lemme know if you want the other one too! :)
**Note: I'm not sure which terminology you're most familiar with using in class but just know that y is the same thing as/another way or writing y(x) or f(x) on the left side of the equation. Similarly, y' = y'(x) = f'(x) = d/dx(y) = dy/dx = d/dx[y(x)] = d/dx[f(x)].**
So we know the First Principles from the Fundamental Theorem of Calculus defines the derivative as a limit:
f'(x) = lim(h→0)⡠[f(x+h)-f(x)]/h
Our function is:
f(x) = 1/[(x^2)+2]
Similarly:
f(x+h) = 1/[((x+h)^2)+2]
Simplify:
f(x+h) = 1/[(x+h)(x+h)+2]
f(x+h) = 1/[(x^2+2xh+h^2+2]
Then we plug it into the limit and simplify:
f'(x) = lim(h→0)⡠〖[1/(((x+h)^2)+2)] - [1/((x^2)+2))]〗/h
f'(x) = lim(h→0)⡠〖[1/(((x+h)^2)+2)] - [1/((x^2)+2))]〗*(1/h)
f'(x) = lim(h→0)⡠〖[1/(((x+h)^2)+2)]*[(x^2)+2]/[(x^2)+2]
- [1/((x^2)+2))]*[((x+h)^2)+2]/[((x+h)^2)+2]〗*(1/h)
f'(x) = lim(h→0)⡠〖[[(x^2)+2]-[((x+h)^2)+2]]/[[(x^2)+2]*[((x+h)^2)+2]]〗*(1/h)
f'(x) = lim(h→0)⡠〖[[(x^2)+2]-[(x^2+2xh+h^2)+2]]/[[(x^2)+2]*[((x^2+2xh+h^2)+2]]〗*(1/h)
f'(x) = lim(h→0)⡠〖[(x^2)+2-(x^2)-2xh-(h^2)-2]/[[(x^2)+2]*[((x^2+2xh+h^2)+2]]〗*(1/h)
f'(x) = lim(h→0)⡠〖[(x^2)+2-(x^2)-2xh-(h^2)-2]/[(x^4)+2(x^3)h+(x^2)(h^2)+2(x^2)+2(x^2)+4xh+2(h^2)+4]〗*(1/h)
f'(x) = lim(h→0)⡠〖[-2xh-(h^2)]/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]〗*(1/h)
f'(x) = lim(h→0)⡠(2h)〖(-x-h)/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]〗*(1/h)
f'(x) = lim(h→0)⡠(2)〖(-x-h)/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]〗
Finally, plug in 0 for h:
f'(x) = lim(h→0)⡠(2)〖(-x-(0))/[(x^4)+2(x^3)(0)+(x^2)((0)^2)+4(x^2)+4x(0)+2((0)^2)+4]〗 f'(x) = (2)[-x/[(x^4)+4(x^2)+4]
Factor the denominator:
f'(x) = (2)[-x/[(x^2)+2)^2]
Final answer: f'(x) = -2x/[((x^2)+2)^2]
Hope this helped!! ~Casey
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