Question

Assuming the derivatives of $\sinh x$ and $\cosh x$, use the quotient rule to prove that is,
$y=\tanh x=\dfrac{\sinh x}{\cosh x}$, then $\dfrac{dy}{dx}={{\operatorname{sech}}^{2}}x$.

Answer 1

Hint: Write the hyperbolic value of $\sinh x$ and $\cosh x$. Find the value of y by substituting the value of $\sinh x$ and $\cosh x$. Find its derivative using quotient rule of differentiation given by, $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ and $\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.

Complete step by step solution:
The hyperbolic sine and hyperbolic cosine functions are given as, $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ and $\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.
We have been given that, $y=\tanh x=\dfrac{\sinh x}{\cosh x}$.
Now substitute the value of $\sinh x$ and $\cosh x$ in the above expression.
$y=\tanh x=\dfrac{\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}}{\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}}$
Let us cancel out the common denominator 2 from the above expression.
$\therefore y=\tanh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$
Let us take, $\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]$
Hence let us solve using the quotient rule which is given as,
$d\left[ \dfrac{f\left( x \right)}{g\left( x \right)} \right]=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x \right)g'\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$
Here, $f\left( x \right)={{e}^{x}}-{{e}^{-x}}$ and $g\left( x \right)={{e}^{x}}+{{e}^{-x}}$

& \therefore f'\left( x \right)={{e}^{x}}-\left( -1 \right){{e}^{-x}}={{e}^{x}}+{{e}^{-x}} \\\ & g'\left( x \right)={{e}^{x}}+\left( -1 \right){{e}^{-x}}={{e}^{x}}-{{e}^{-x}} \\\ \end{aligned}$$ Let us substitute these values in the formula of the quotient rule. $$\dfrac{dy}{dx}=\dfrac{\left( {{e}^{x}}+{{e}^{-x}} \right)\left( {{e}^{x}}+{{e}^{-x}} \right)-\left( {{e}^{x}}-{{e}^{-x}} \right)\left( {{e}^{x}}-{{e}^{-x}} \right)}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}$$ Let us open brackets and simplify it, $$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{\left[ {{e}^{2x}}+{{e}^{x-x}}+{{e}^{x-x}}+{{e}^{-2x}} \right]-\left[ {{e}^{2x}}-{{e}^{x-x}}-{{e}^{x-x}}+{{e}^{-2x}} \right]}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}} \\\ & \dfrac{dy}{dx}=\dfrac{{{e}^{2x}}+{{e}^{0}}+{{e}^{0}}+{{e}^{-2x}}-{{e}^{-2x}}+{{e}^{0}}+{{e}^{0}}-{{e}^{-2x}}}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}=\dfrac{4{{e}^{0}}}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}} \\\ \end{aligned}$$ We know that, $${{e}^{0}}=1$$. $$\therefore \dfrac{dy}{dx}=\dfrac{4}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}$$ - (1) We know that, $$\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$$ $$\therefore {{\cos }^{2}}hx={{\left[ \dfrac{\left( {{e}^{x}}+{{e}^{-x}} \right)}{2} \right]}^{2}}=\dfrac{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}{4}$$ We know that, $$\operatorname{sech}x=\dfrac{1}{\cosh x}$$. $$\therefore {{\sec }^{2}}hx=\dfrac{1}{{{\cos }^{2}}hx}=\dfrac{1}{\dfrac{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}{4}}=\dfrac{4}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}$$ $$\therefore {{\sec }^{2}}hx=\dfrac{4}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}$$ - (2) Comparing (1) and (2) we can say that, $$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{4}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}={{\sec }^{2}}hx \\\ & \therefore \dfrac{dy}{dx}={{\sec }^{2}}hx \\\ \end{aligned}$$ Hence we proved the required. Note: The derivative of $$\tan x$$ is $${{\sec }^{2}}x$$. Thus the derivative of $$\tanh x$$ is equal to $${{\operatorname{sech}}^{2}}x$$. $$\dfrac{d}{dx}\tan x={{\sec }^{2}}x$$ $$\dfrac{d}{dx}\tanh x={{\sec }^{2}}hx$$ There are lots of terms involved while applying the quotient rule and care must be taken during its simplification. Any mistake will lead to loss time as we will not be able to prove the results and leave the question as it is.