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What is the derivative of f plus g of 3 and f times g of 3 given that f of 3 equals 5 d dx f of 3 equals 1.1 g of 3 equals -4 d dx g of 3 equals 7 Also please explain QUICK THANK YOU?
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
Mean and standard deviation of probability density function?
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).
What does fyd mean on x-factor?
F = For Y = Your D = Descrition i think lol
What is the derivative of 3 to the power of 2 multiplied by f of x?
I think you mean; f(x) = 3x^2 d/dx(3x^2) = 6x
What is the first derivative of the quantity 1 plus x negative squared?
!f you mean d/dx (1 + x-2) = -2x-3 If you mean d/dx (1 + x)-2 = -2(1 + x)-3
Where can you find freak the freak out notes for the piano?
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!
How to prove that differentiating in the space of smooth functions is a linear transformation?
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.
What is the equation for this A set of 7 numbers has a mean of 9 What additional number must be included in this set to create a new set with a mean that is 3 less than the mean of the original set?
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).
How do you differentiate sine x squared?
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)
How do you differentiate a fraction with x as a numerator?
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)
What are the zeros of f of x plus 2?
I think you might mean f(x)+2? Or do you mean f(x+2)? Either way it depends on what f(x) is.
What is the first derivative of modulus function?
f(x)=|x| |x| = sqrt(x^2) f(x) = sqrt (x^2) f'(x)=1/2sqrt(x^2) d/dx(x^2) f'(x) = 2x /2sqrt(x^2) f'(x) = x/sqrt(x^2) f'(x)= x /|x| --------------------- Note: d/dx [sqrt(x^2] = d/dx [x^(1/2)] =(1/2)x^(-1/2)=1/2x^(1/2)=1/2sqrt(x)
