New Notes Blog: `f(x) = cosh(8x+1)` Find the derivative of the function

Monday, 28 October 2013

$\displaystyle{f{{\left({x}\right)}}}={\cosh{{\left({8}{x}+{1}\right)}}}$ Find the derivative of the function

$\displaystyle{f{{\left({x}\right)}}}={\cosh{{\left({8}{x}+{1}\right)}}}$

Take note that the derivative formula of cosh is

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\cosh{{\left({u}\right)}}}\right]}={\sinh{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying this formula, the derivative of the function will be

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{\left.{d}{x}}}{\left[{\cosh{{\left({8}{x}+{1}\right)}}}\right]}$

$\displaystyle{f{'}}{\left({x}\right)}={\sinh{{\left({8}{x}+{1}\right)}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({8}{x}+{1}\right)}$

$\displaystyle{f{'}}{\left({x}\right)}={\sinh{{\left({8}{x}+{1}\right)}}}\cdot{8}$

$\displaystyle{f{'}}{\left({x}\right)}={8}{\sinh{{\left({8}{x}+{1}\right)}}}$

Therefore, the derivative of the function is $\displaystyle{f{'}}{\left({x}\right)}={8}{\sinh{{\left({8}{x}+{1}\right)}}}$ .

$\displaystyle{f{{\left({x}\right)}}}={\cosh{{\left({8}{x}+{1}\right)}}}$

Take note that the derivative formula of cosh is

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\cosh{{\left({u}\right)}}}\right]}={\sinh{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying this formula, the derivative of the function will be

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{\left.{d}{x}}}{\left[{\cosh{{\left({8}{x}+{1}\right)}}}\right]}$

$\displaystyle{f{'}}{\left({x}\right)}={\sinh{{\left({8}{x}+{1}\right)}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({8}{x}+{1}\right)}$

$\displaystyle{f{'}}{\left({x}\right)}={\sinh{{\left({8}{x}+{1}\right)}}}\cdot{8}$

$\displaystyle{f{'}}{\left({x}\right)}={8}{\sinh{{\left({8}{x}+{1}\right)}}}$

Therefore, the derivative of the function is $\displaystyle{f{'}}{\left({x}\right)}={8}{\sinh{{\left({8}{x}+{1}\right)}}}$ .

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