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Why does a number get smaller when you multiply by a number less than one?
Because the number is decreasing in size (i.e. less than one) which means you are making it smaller. Try dividing a big number with any fraction. You'll see that it decreases. This is because a fraction is smaller than one and so it will decrease. Another way of looking at it is if you multiply a number by 1, you get the same number. Multiply by 2 and you get twice the number. Therefore, multiplying a number by less than one and your bound to get a smaller number.
What is x plus y to the second power plus z to the second power?
(x+y)^2+z^2=x^2+y^2+z^2+2xy or ((x+y)^2+z)^2= (x^2+y^2+2xy+z)^2= x^4+y^4+z^2+6x^2y^2+4x^3y+2x^2z^2+4xy^3+4xyz^2+2z^2y^2
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)]/hOur 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))]〗/hf'(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
What is integral tan3x sec2x?
-(4*log(2*cos(4*x)-4*cos(2*x)+3)-3*log(2*cos(4*x)+2)-2*log(2*cos(2*x)+2))/12
What does 2x1 equal?
2
