Question

What is the derivative of \[\cosh x\]?

Answer 1

Hint: To solve this question we have to use a formula which converts \[\cosh x\] into \[{e^x}\]. If we convert \[\cosh x\]into \[{e^x}\] then the derivative of that function is too easy. If we try to find the derivative of \[\cosh x\] directly then we are unable to find the derivative of \[\cosh x\]. The formula of \[\cosh x\]in terms of \[{e^x}\] is. After differentiating, convert the equation into the hyperbolic or trigonometric. That equation is converted into \[\sinh x\]
\[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\]

Complete step-by-step solution:
Given;
A trigonometric function that is
\[f(x) = \cosh x\]
To find,
Derivative of that function
Formula used:
Formula for converting \[\cosh x\] to \[\dfrac{{{e^x} + {e^{ - x}}}}{2}\]
\[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\]
And formula for converting \[\dfrac{{{e^x} - {e^{ - x}}}}{2}\] to \[\sinh x\].
The given function is
 \[f(x) = \cosh x\] ……………………………(i)
\[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\] ……(ii)
From equation (i) and equation (ii)
\[f(x) = \dfrac{{{e^x} + {e^{ - x}}}}{2}\]
Now we have to find the derivative of function \[f(x)\]
Differentiating both side with respect to \[x\]
\[\dfrac{{d(f(x))}}{{dx}} = \dfrac{{d\dfrac{{{e^x} + {e^{ - x}}}}{2}}}{{dx}}\]
Taking 2 outside the derivative because 2 is constant and the constant part is taken outside from the derivative.
\[\dfrac{{d(f(x))}}{{dx}} = \dfrac{1}{2}\dfrac{{d({e^x} + {e^{ - x}})}}{{dx}}\]
Put the value of \[f(x)\] from equation (i)
\[\dfrac{{d\cosh x}}{{dx}} = \dfrac{1}{2}\dfrac{{d({e^x} + {e^{ - x}})}}{{dx}}\]
Using the distributive property of derivative
\[\dfrac{{d\cosh x}}{{dx}} = \dfrac{1}{2}(\dfrac{{d({e^x})}}{{dx}} + \dfrac{{d({e^{ - x}})}}{{dx}})\]
Derivative of \[{e^x}\]is \[{e^x}\]
Using the chain rule of derivative
\[\dfrac{{d\cosh x}}{{dx}} = \dfrac{1}{2}({e^x} - {e^{ - x}})\] ……………(iii)
As, we know
\[\sinh x = \dfrac{1}{2}({e^x} - {e^{ - x}})\] …………………(iv)
Putting the value from equation (iv) to equation (iii)
\[\dfrac{{d\cosh x}}{{dx}} = \sinh x\]
Final answer:
Derivative of \[\cosh x\] is
\[ \Rightarrow \dfrac{{d\cosh x}}{{dx}} = \sinh x\]

Note: To solve these types of questions we must know all the formulas of hyperbolic trigonometry. Without that formula we are unable to solve the derivative of that function. At last we have to convert the last expression of \[{e^x}\] into the hyperbolic trigonometric. In this particular case we first change \[\cosh x\] to \[\dfrac{{{e^x} + {e^{ - x}}}}{2}\]and after solving \[\dfrac{{{e^x} + {e^{ - x}}}}{2}\] we get different expression like \[\dfrac{{{e^x} - {e^{ - x}}}}{2}\] and then again convert that to \[\sinh x\].