Question

Prove the formulas given in this table for the derivatives of the functions cosh, tanh, csch, sech, and coth. Which of the following are proven correctly? (Select all that apply.)

\(\square \frac{d}{d x}(\operatorname{coth} x)=\frac{d}{d x}\left(\frac{\sinh x}{\cosh x}\right)=\frac{\cosh x \cosh x-\sinh x \sinh x}{\cosh ^{2} x}=\frac{\cosh ^{2} x-\sinh ^{2} x}{\cosh ^{2} x}=-\frac{1}{\cosh ^{2} x}=-\operatorname{csch}^{2} x\) \(\square \frac{d}{d x}(\operatorname{csch} x)=\frac{d}{d x}\left(\frac{1}{\sinh x}\right)=-\frac{\cosh x}{\sinh ^{2} x}=-\frac{1}{\sinh x} \cdot \frac{\cosh x}{\sinh x}=-\operatorname{csch} x \operatorname{coth} x\)

\(\square \frac{d}{d x}(\cosh x)=\frac{d}{d x}\left[\frac{1}{2}\left(e^{x}-e^{-x}\right)\right]=\frac{1}{2}\left(e^{x}+e^{-x}\right)=\sinh x\)

\(\square \frac{d}{d x}(\operatorname{csch} x)=\frac{d}{d x}\left(\frac{1}{\sinh x}\right)=-\frac{\cosh ^{2} x}{\sinh ^{2} x}=-\frac{1}{\sinh x} \cdot \frac{\cosh ^{2} x}{\sinh x}=-\operatorname{csch} x \operatorname{coth} x\)

\(\square \frac{d}{d x}(\operatorname{sech} x)=\frac{d}{d x}\left(\frac{1}{\cosh x}\right)=-\frac{\sinh x}{\cosh ^{2} x}=-\frac{1}{\cosh x} \cdot \frac{\sinh x}{\cosh x}=-\operatorname{sech} x \tanh x\)

Answer 1

Proving the formula \(\frac{d}{d x}(\operatorname{coth} x)=-\operatorname{csch}^{2} x\)

$$ \begin{aligned} &\frac{d}{d x}(\operatorname{coth} x)=\frac{d}{d x}\left(\frac{\cosh x}{\sinh x}\right) \\ &\Rightarrow \frac{d}{d x}(\operatorname{coth} x)=\left(\frac{\left.\sinh x \frac{d}{d x} \cosh x-\cosh x \frac{d}{d x} \sinh x\right)}{\sinh ^{2} x}\right) \quad[\text { By quotient rule of differentiation }] \\ &\Rightarrow \frac{d}{d x}(\operatorname{coth} x)=\left(\frac{\sinh x \cdot \sinh x-\cosh x \cdot \cosh x}{\sinh ^{2} x}\right) \quad\left[\operatorname{since} \frac{d}{d x} \sinh x=\cosh x, \frac{d}{d x} \cosh x=\sinh x\right] \\ &\Rightarrow \frac{d}{d x}(\operatorname{coth} x)=\left(\frac{\sinh ^{2} x-\cosh ^{2} x}{\sinh ^{2} x}\right) \\ &\Rightarrow \frac{d}{d x}(\operatorname{coth} x)=\left(\frac{-1}{\sinh ^{2} x}\right) \quad\left[\operatorname{sincecosh}^{2} x-\sinh ^{2} x=1\right] \\ &\Rightarrow \frac{d}{d x}(\operatorname{coth} x)=-\operatorname{csch}^{2} x \end{aligned} $$

Hence the given prove is wrong as in the given prove it is used that \(\left(\operatorname{coth} x=\frac{\sinh x}{\cosh x}\right)\).

Proving the formula \(\frac{d}{d x}(\operatorname{csch} x)=-\operatorname{csch} x \operatorname{coth} x\),

$$ \begin{aligned} &\frac{d}{d x}(\operatorname{csch} x)=\frac{d}{d x}\left(\frac{1}{\sinh x}\right) \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=\left(\frac{-\frac{d}{d x} \sinh x}{\sinh ^{2} x}\right) \quad[\text { By Quotient Rule of Differentiation }] \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=\left(\frac{-\cosh x}{\sinh ^{2} x}\right)\left[\operatorname{since} \frac{d}{d x} \sinh x=\cosh x\right] \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=-\frac{1}{\sinh x} \frac{\cosh x}{\sinh x} \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=-\operatorname{csch} x \operatorname{coth} x \end{aligned} $$

Hence the given prove is correct.

Proving the formula \(\frac{d}{d x}(\cosh x)=\sinh x\),

$$ \begin{aligned} &\frac{d}{d x}(\cosh x)=\frac{d}{d x}\left(\frac{1}{2}\left(e^{x}-e^{-x}\right)\right) \quad\left[\because \cosh x=\frac{e^{x}-e^{-x}}{2}\right] \\ &\Rightarrow \frac{d}{d x}(\cosh x)=\frac{1}{2}\left(\frac{d}{d x} e^{x}-\frac{d}{d x} e^{-x}\right) \\ &\Rightarrow \frac{d}{d x}(\cosh x)=\frac{1}{2}\left(e^{x}+e^{-x}\right) \\ &\Rightarrow \frac{d}{d x}(\cosh x)=\sinh x \quad\left[\because \sinh x=\frac{e^{x}+e^{-x}}{2}\right] \end{aligned} $$

Hence the given prove is correct.

Proving the formula \(\frac{d}{d x}(\operatorname{csch} x)=-\operatorname{csch} x \operatorname{coth} x\)

$$ \begin{aligned} &\frac{d}{d x}(\operatorname{csch} x)=\frac{d}{d x}\left(\frac{1}{\sinh x}\right) \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=\left(\frac{-\frac{d}{d x} \sinh x}{\sinh ^{2} x}\right) \quad[\text { By quotient rule of differentiation }] \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=\left(\frac{-\cosh x}{\sinh ^{2} x}\right)\left[\operatorname{since} \frac{d}{d x} \sinh x=\cosh x, \frac{d}{d x} \cosh x=\sinh x\right] \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=-\frac{1}{\sinh x} \frac{\cosh x}{\sinh x} \\ &\Rightarrow \frac{d}{d x}(\operatorname{csch} x)=-\operatorname{csch} x \operatorname{coth} x \end{aligned} $$

Hence the given prove is wrong as in the given prove it is used that \(\left(\frac{d}{d x} \sinh x=\cosh ^{2} x\right)\).

Proving the formula \(\frac{d}{d x}(\operatorname{sech} x)=-\operatorname{sech} x \tanh x\),

$$ \begin{aligned} &\frac{d}{d x}(\operatorname{sech} x)=\frac{d}{d x}\left(\frac{1}{\cosh x}\right) \\ &\Rightarrow \frac{d}{d x}(\operatorname{sech} x)=\left(\frac{-\frac{d}{d x} \cosh x}{\cosh ^{2} x}\right) \quad \text { [By Quotient Rule of Dif } \\ &\Rightarrow \frac{d}{d x}(\operatorname{sech} x)=\left(\frac{-\sinh x}{\cosh ^{2} x}\right) \quad\left[\text { Since } \frac{d}{d x} \cosh x=\sinh x\right] \\ &\Rightarrow \frac{d}{d x}(\operatorname{sech} x)=-\frac{1}{\cosh x} \frac{\sinh x}{\cosh x} \\ &\Rightarrow \frac{d}{d x}(\operatorname{sech} x)=-\operatorname{sech} x \tanh x \end{aligned} $$

Hence the given prove is correct.

Answers can be found in the Explanation section.