d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)] or f'(x) + g'(x) when x = 3, d/dx [f(x) + g(x)] = f'(3) + g'(3) = 1.1 + 7 = 8.1 d/dx [f(x)*g(x)] = f(x)*d/dx[g(x)] + d/dx[f(x)]*g(x) when x = 3, d/dx [f(x)*g(x)] = f(3)*g'(3) + f'(3)*g(3) = 5*7 + 1.1*(-4) = 35 - 4.4 = 31.1
Suppose the probability density function is f(x), defined over a domain D Then the mean is E(X) = x*f(x) integrated with respect to x over D. Calculate E(X2) = x2*f(x) integrated with respect to x over D. Then Variance(X) = E(X2) - [E(X)]2 and Standard Deviation = sqrt(Variance).
F = For Y = Your D = Descrition i think lol
I think you mean; f(x) = 3x^2 d/dx(3x^2) = 6x
I personally went through a YouTube video of it and wrote down the notes in A-G. Hold down the notes in Bold.F x 8C x 8D x 8B# x 7D,D,D,D,F,D,D,F,D,D,F,D,D,D,D,C,D,D,C,D,D,CF,F,E,E,D,C-D F,F,E,E,D,C-DF-F,F,F,G,F,D,C, F,F,F C,E,G,F,A F,A A,A#,A E,G G,F,G,A,G C,F,F,C B# F,G,FThat's all I've gotten up to!Hope this helps! It sounds great once you get to play it over and over!
!f you mean d/dx (1 + x-2) = -2x-3 If you mean d/dx (1 + x)-2 = -2(1 + x)-3
Recall that a linear transformation T:U-->V is one such that 1) T(x+y)=T(x)+T(y) for any x,y in U 2) T(cx)=cT(x) for x in U and c in R All you need to do is show that differentiation has these two properties, where the domain is C^(infinity). We shall consider smooth functions from R to R for simplicity, but the argument is analogous for functions from R^n to R^m. Let D by the differential operator. D[(f+g)(x)] = [d/dx](f+g)(x) = lim(h-->0)[(f+g)(x+h)-(f+g)(x)]/h = lim(h-->0)[f(x+h)+g(x+g)-f(x)-g(x)]/h (since (f+g)(x) is taken to mean f(x)+g(x)) =lim(h-->0)[f(x+h)-f(x)]/h + lim(h-->0)[g(x+h) - g(x)]/h since the sum of limits is the limit of the sums =[d/dx]f(x) + [d/dx]g(x) = D[f(x)] + D[g(x)]. As for ths second criterion, D[(cf)(x)]=lim(h-->0)[(cf)(x+h)-(cf)(x)]/h =lim(h-->0)[c[f(x+h)]-c[f(x)]]/h since (cf)(x) is taken to mean c[f(x)] =c[lim(h-->0)[f(x+h)-f(x)]/h] = c[d/dx]f(x) = cD[f(x)]. since constants can be factored out of limits. Therefore the two criteria hold, and if you wished to prove this for the general case, you would simply apply the same procedure to the Jacobian matrices corresponding to Df.
Call the specified numbers a, b, c, d, e, f, and g and the unknown number x. From the problem statement, (a + b + c + d + e + f + g)/7 = 9 [from the definition of "mean" of a set of numbers]. Also (a + b + c + d + e + f + g + x)/(7 + 1) = 9 - 3 = 6, or x/8 = 6 - [(a + b + c + d + e + f + g)/8], or x = 48 - (a + b + c + d + e + f + g).
Suppose you wish to differentiate x/f(x) where f(x) is a differentiable function of x, and writing f for f(x) and f'(x) for the derivative of f(x), d/dx (x/f) = [f - x*f']/(f2)
Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)
I think you might mean f(x)+2? Or do you mean f(x+2)? Either way it depends on what f(x) is.
ex+f = c -dx ex+dx = c -f x(e+d) = c -f x = c -f/(e+d)