Can someone show me how to do this?
P phillyx New member Joined Sep 30, 2009 Messages 1 Sep 30, 2009 #1 Can someone show me how to do this? Attachments calc.JPG 5.5 KB · Views: 69
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Oct 1, 2009 #2 f(x) = xcosh−1(x4) −(x2−16)\displaystyle f(x) \ = \ xcosh^{-1}\bigg(\frac{x}{4}\bigg) \ -\sqrt(x^{2}-16)f(x) = xcosh−1(4x) −(x2−16) f ′ (x) = (1)cosh−1(x4)+x1/4((x2/16)−1)−12(x2−16)−1/2(2x)\displaystyle f \ ' \ (x) \ = \ (1)cosh^{-1}\bigg(\frac{x}{4}\bigg)+x \frac{1/4}{\sqrt((x^{2}/16)-1)} -\frac{1}{2}(x^{2}-16)^{-1/2}(2x)f ′ (x) = (1)cosh−1(4x)+x((x2/16)−1)1/4−21(x2−16)−1/2(2x) \(\displaystyle = \ cosh^-1}\bigg(\frac{x}{4}\bigg)+\frac{x}{\sqrt(x^{2}-16)}-\frac{x}{\sqrt(x^{2}-16)}\) = cosh−1(x4)\displaystyle = \ cosh^{-1}\bigg(\frac{x}{4}\bigg)= cosh−1(4x) Hence f ′ (6) = cosh−1(32) = ln(3+52). QED\displaystyle Hence \ f \ ' \ (6) \ = \ cosh^{-1}\bigg(\frac{3}{2}\bigg) \ = \ ln\bigg(\frac{3+\sqrt 5}{2}\bigg). \ QEDHence f ′ (6) = cosh−1(23) = ln(23+5). QED
f(x) = xcosh−1(x4) −(x2−16)\displaystyle f(x) \ = \ xcosh^{-1}\bigg(\frac{x}{4}\bigg) \ -\sqrt(x^{2}-16)f(x) = xcosh−1(4x) −(x2−16) f ′ (x) = (1)cosh−1(x4)+x1/4((x2/16)−1)−12(x2−16)−1/2(2x)\displaystyle f \ ' \ (x) \ = \ (1)cosh^{-1}\bigg(\frac{x}{4}\bigg)+x \frac{1/4}{\sqrt((x^{2}/16)-1)} -\frac{1}{2}(x^{2}-16)^{-1/2}(2x)f ′ (x) = (1)cosh−1(4x)+x((x2/16)−1)1/4−21(x2−16)−1/2(2x) \(\displaystyle = \ cosh^-1}\bigg(\frac{x}{4}\bigg)+\frac{x}{\sqrt(x^{2}-16)}-\frac{x}{\sqrt(x^{2}-16)}\) = cosh−1(x4)\displaystyle = \ cosh^{-1}\bigg(\frac{x}{4}\bigg)= cosh−1(4x) Hence f ′ (6) = cosh−1(32) = ln(3+52). QED\displaystyle Hence \ f \ ' \ (6) \ = \ cosh^{-1}\bigg(\frac{3}{2}\bigg) \ = \ ln\bigg(\frac{3+\sqrt 5}{2}\bigg). \ QEDHence f ′ (6) = cosh−1(23) = ln(23+5). QED
