Find a formula for F^n (x) if F(x) = Ln(x-1). Prove formula by using induction?
Hi f(x) = ln(x - 1) f'(x) = 1/(x - 1) f''(x) = -1/(x - 1)^2 f'''(x) = 2/(x - 1)^3 … f^n(x) = (-1)^(n - 1)*(n - 1)!/(x - 1)^n We must show that this is true for n = 1: f^1(x) = (-1)^(1 - 1)*(1 - 1)!/(x - 1)^1 = 1/(x - 1) Now, show that if it's true for f^k(x), it's true for f^(k + 1)(x). If f^k(x) = (-1)^(k - 1)*(k - 1)!/(x - 1)^k, then we have: f^(k + 1)(x) = [f^k(x)]' = [(-1)^(k - 1)*(k - 1)!/(x - 1)^k]' = (-1)^(k - 1)*(k - 1)!*(-k)/(x - 1)^(k + 1) = (-1)^k*k!/(x - 1)^(k + 1) Which agrees with the formula: f^(k + 1)(x) = (-1)^(k + 1 - 1)*(k + 1 - 1)!/(x - 1)^(k + 1) = (-1)^k*k!/(x - 1)^(k + 1) Since we have shown that f^k(x) = (-1)^(k - 1)*(k - 1)!/(x - 1)^k = [f^k(x)]' implies that f^(k + 1)(x) = (-1)^k*k!/(x - 1)^(k + 1) and the formula works for k = 1, the formula f^n(x) = (-1)^(n - 1)*(n - 1)!/(x - 1)^n must be true. I hope this helps!
Find a formula for RN for the given function and interval. Then compute the area under the graph as a limit.?
What is RN? Let y=f(x)≥0 for x∈[a,b]. Then the area of the region between the graph of the function and the x-axis over the specified interval is: Area=lim[n-->inf] [(b-a)/n]∑[i=1 to n] f[a+i(b-a)/n] a=0, b=1 Area=lim[n-->inf] [(1-0)/n]∑[i=1 to n] f[0+i(1-0)/n]= lim[n-->inf] (1/n)∑[i=1 to n] (i/n)²= lim[n-->inf] (1/n³)∑[i=1 to n] i²= lim[n-->inf] (1/n³)*n(n+1)(2n+1)/6=2/6=1/3
Find the formula for the nth derivative of f?
f(x)=xe^7x => f´(x) = 1*e^(7x) + 7x*e^(7x) = (1 + 7x)*e^(7x) = (1*7^0 + 7^1*x)*e^(7x) => f´´(x) = 7*e^(7x) + (1 +7x)*7*e^(7x) = (14 + 49x)*e^(7x) = (2*7^1 + 7^2*x)*e^(7x) => f´´´(x) = 49*e^(7x) + (14 + 49x)*7*e^(7x) = (147 + 343*x)*e^(7x) = (3*7^2 + 7^3*x)*e^(7x) => f^(n)(x) = (n*7^(n-1) + 7^n*x)*e^(7x) ============================ You can proof this by mathematical induction!
How do I find the general formula for f^n(x)?
Through using the link provided by Gianlino or otherwise, it's easy to verify that f^(n)(x) = n * [(n-1)-th derivative of sin(ax)] + 1 * [n-th derivative of sin(ax)]. (Just write out the first few derivatives of f without simplifying the derivatives of sin(ax).) Now the crux of the problem: What is the k-th derivative of sin x? It's easy to check that the derivatives of sin x cycle in 4: cos x, -sin x, -cos x, and finally back to sin x. To capture this all at once, I claim that (d^k/dx^k) (sin x) = sin(x + πk/2) for any integer k ≥ 0. Note that (using the addition formulas for sine and cosine) k = 0 ==> sin(x + 0) = sin x k = 1 ==> sin(x + π/2) = cos x k = 2 ==> sin(x + π) = -sin x k = 3 ==> sin(x + 3π/2) = -cos x k = 4 ==> sin(x + 2π) = sin x ... (No need to go further since we can use the periodicity of sine and cosine for higher values of k.) ------------------ Armed with this result and the Chain Rule (to deal with the constant 'a'): f^(n)(x) = na^(n-1) sin(ax + π(n-1)/2) + a^n sin(ax + πn/2). I hope this helps!
Finding a Formula for the Nth derivative....?
Let f (x) = x*(e^8x) Find a formula for the nth derivative of f, where n is any positive integer. Use x and n in your answer if needed. I know that the formula starts out with: (8^n)x(e^(8x))+... but I don't know how to figure out the second part. Can anyone please help me?
Finding the value of n in the formula?
In an experiment to fins the number of molecules of water of crystallisation in sodium sulfate crystals.Na2SO4*nH2O,3.22g of sodium sulfate crystals were heated gently.when all the water crystallisation had been driven off,1.42g of anhydrous sodium sulfate was left.Find the value for n in the formula The * represents addition
Find a formula for the nth derivative of f, where n is any positive integer (Calculus 2)?
Write out the first few to find a pattern. f '(x) = e^(8x) + x * 8e^(8x) = (1 + 8x) e^(8x). f ''(x) = 8 e^(8x) + (1 + 8x) * 8e^(8x) .......= (2 * 8 + 8^2 x) e^(8x) .......= 8(2 + 8x) e^(8x). f '''(x) = 8 * 8 e^(8x) + 8(2 + 8x) * 8e^(8x) ........= [8 * 8 + 8(2 + 8x) * 8] e^(8x) ........= 8^2 [1 + (2 + 8x)] e^(8x) ........= 8^2 (3 + 8x) e^(8x) Pattern: f^(n)(x) = 8^(n-1) (n + 8x) e^(8x). I hope this helps!