Differentiate using the chain rule.
\begin{align*} f'(x) & = \cosh\left(x^{4/3}\right)\cdot \frac d {dx}\left(x^{4/3}\right)\\[6pt] & = \cosh\left(x^{4/3}\right)\cdot \frac 4 3 x^{1/3} \end{align*}
Simplify.
$$f'(x) = \frac 4 3 x^{1/3}\cosh\left(x^{4/3}\right)$$
$$\displaystyle f'(x) = \frac 4 3 x^{1/3}\cosh\left(x^{4/3}\right)$$
Differentiate using the chain rule.
$$ \begin{align*} f'(x) & = \sinh\left(3-6x^7\right)\cdot \frac d {dx}\left(3-6x^7\right)\\[6pt] & = \sinh\left(3-6x^7\right)\cdot (-42x^6)\\[6pt] & = -42x^6\sinh\left(3-6x^7\right) \end{align*} $$
$$\displaystyle f'(x) = -42x^6\sinh\left(3-6x^7\right)$$
Rewrite the function so the square-root is expressed in exponent form.
$$ f(x) = \tanh\left(x^{1/2}\right) $$
Differentiate using the chain rule.
$$ \begin{align*} f'(x) & = \sech^2\left(x^{1/2}\right)\cdot \frac d {dx}\left(x^{1/2}\right)\\[6pt] & = \sech^2\left(x^{1/2}\right)\cdot \frac 1 2 x^{-1/2} \end{align*} $$
Simplify.
$$ \begin{align*} f'(x) & = \sech^2\left(x^{1/2}\right)\cdot \frac 1 2 x^{-1/2}\\[6pt] & = \sech^2\left(x^{1/2}\right)\cdot \frac 1 2 \cdot\frac 1 {x^{1/2}}\\[6pt] & = \sech^2\left(\sqrt x\right)\cdot \frac 1 {2\sqrt x}\\[6pt] & = \frac{\sech^2\left(\sqrt x\right)}{2\sqrt x} \end{align*} $$
(Optional) Rationalize the denominator.
$$ f'(x) = \frac{\sqrt x\,\sech^2\left(\sqrt x\right)}{2x} $$
$$ \displaystyle f'(x) = \frac{\sqrt x\,\sech^2\left(\sqrt x\right)}{2x} $$
Differentiate using the chain rule.
$$ \begin{align*} \frac d {dx}\left(\coth 7x^5\right) & = -\csch^2 7x^5 \cdot \frac d {dx}\left(7x^5\right)\\[6pt] & = -\csch^2 7x^5 \cdot 35x^4\\[6pt] & = -35x^4\csch^2 7x^5 \end{align*} $$
$$\displaystyle \frac d {dx}\left(\coth 7x^5\right) = -35x^4\csch^2 7x^5$$
Rewrite the function so the $$\frac 1 x$$ is in exponent form.
$$ \frac d {dx}\left[\sech\left(\frac 1 x\right)\right] = \frac d {dx}\left[\sech\left(x^{-1}\right)\right] $$
Differentiate using the chain rule.
$$ \begin{align*} \frac d {dx}\left[\sech\left(x^{-1}\right)\right] & = -\sech\left(x^{-1}\right)\tanh\left(x^{-1}\right)\cdot \frac d {dx}\left(x^{-1}\right)\\[6pt] & = -\sech\left(x^{-1}\right)\tanh\left(x^{-1}\right)\cdot \left(-x^{-2}\right) \end{align*} $$
Simplify.
$$ \begin{align*} \frac d {dx}\left[\sech\left(x^{-1}\right)\right] & = -\sech\left(x^{-1}\right)\tanh\left(x^{-1}\right)\cdot \left(-x^{-2}\right)\\[6pt] & = x^{-2}\sech\left(x^{-1}\right)\tanh\left(x^{-1}\right)\\[6pt] & = \frac 1 {x^2}\sech\left(\frac 1 x\right)\tanh\left(\frac 1 x\right) \end{align*} $$
$$\displaystyle \frac d {dx}\left[\sech\left(\frac 1 x\right)\right] = \frac 1 {x^2}\sech\left(\frac 1 x\right)\tanh\left(\frac 1 x\right)$$
Differentiate using the chain rule.
$$ \begin{align*} f'(x) & = -\csch\left(\frac \pi 2 x^2\right)\coth\left(\frac \pi 2 x^2\right)\cdot \frac d {dx}\left(\frac \pi 2 x^2\right)\\[6pt] & = -\csch\left(\frac \pi 2 x^2\right)\coth\left(\frac \pi 2 x^2\right)\cdot \left(\frac \pi 2 \cdot 2x\right)\\[6pt] & = -\csch\left(\frac \pi 2 x^2\right)\coth\left(\frac \pi 2 x^2\right)\cdot \left(\pi x\right)\\[6pt] & = -\pi x\csch\left(\frac \pi 2 x^2\right)\coth\left(\frac \pi 2 x^2\right) \end{align*} $$
$$ f'(x) = -\pi x\csch\left(\frac \pi 2 x^2\right)\coth\left(\frac \pi 2 x^2\right) $$
Identify the factors in the function.
$$ f(x) = \blue{\csch 2x}\,\red{\coth 2x} $$
Differentiate using the product rule .
$$ \begin{align*} f'(x) & = \blue{-2\csch 2x \coth 2x }\coth 2x + \csch 2x \red{(-2\csch^2 2x)}\\[6pt] & = -2\csch 2x \coth 2x \coth 2x -2\csch 2x \csch^2 2x \\[6pt] & = -2\csch 2x \coth^2 2x -2\csch^3 2x \end{align*} $$
Simplify by factoring.
$$ \begin{align*} f'(x) & = \blue{-2}\csch 2x \coth^2 2x \blue{-2}\csch^3 2x\\[6pt] & = \blue{-2}\left(\red{\csch 2x}\coth^2 2x + \red{\csch^3 2x}\right)\\[6pt] & = -2\red{\csch 2x}\left(\coth^2 2x + \csch^2 2x\right)\\[6pt] & = -2\csch 2x\left(\csch^2 2x - 1 + \csch^2 2x\right)\\[6pt] & = -2\csch 2x\left(2\csch^2 2x - 1\right) \end{align*} $$
$$f'(x) = -2\csch 2x\left(2\csch^2 2x - 1\right)$$
Identify the factors in the function.
$$ f(x) = \blue{\sech \pi x}\,\red{\tanh \pi x} $$
Differentiate using the product rule .
$$ \begin{align*} f'(x) & = \blue{-\pi\sech \pi x\tanh \pi x}\tanh \pi x + \sech \pi x \cdot \red{\pi \sech^2 \pi x}\\[6pt] & = -\pi\sech \pi x\tanh^2 \pi x + \pi\sech^3 \pi x \end{align*} $$
Simplify by factoring.
$$ \begin{align*} f'(x) & = \blue{-\pi}\sech \pi x\tanh^2 \pi x + \blue \pi\sech^3 \pi x\\[6pt] & = \blue{-\pi}\left(\red{\sech \pi x} \tanh^2 \pi x -\red{\sech^3 \pi x}\right)\\[6pt] & = -\pi\,\red{\sech \pi x}\left( \tanh^2 \pi x -\sech^2 \pi x\right)\\[6pt] & = -\pi\,\sech \pi x\left( 1 - \sech^2 \pi x -\sech^2 \pi x\right)\\[6pt] & = -\pi\,\sech \pi x\left(1 - 2\sech^2 \pi x\right) \end{align*} $$
$$\displaystyle f'(x) = -\pi\,\sech \pi x\left(1 - 2\sech^2 \pi x\right)$$
Differentiate using the quotient rule. The parts in $$\blue{blue}$$ are related to the numerator.
$$ \begin{align*} \frac d {dx}\left(\frac{\sinh x} x\right) & = \frac{x\,\blue{\cosh x} - \blue{\sinh x}\cdot 1}{x^2}\\[6pt] & = \frac{x\cosh x - \sinh x}{x^2} \end{align*} $$
$$ \displaystyle \frac d {dx}\left(\frac{\sinh x} x\right) = \frac{x\cosh x - \sinh x}{x^2} $$
Differentiate using the quotient rule. The parts in $$\blue{blue}$$ are related to the numerator.
$$ \begin{align*} f'(x) & = \frac{x\blue{(-\sinh x)} - \blue{(1-\cosh x)}\cdot 1}{x^2}\\[6pt] & = \frac{-x\sinh x - 1+\cosh x}{x^2}\\[6pt] & = \frac{\cosh x - x\sinh x - 1}{x^2} \end{align*} $$
$$ \displaystyle f'(x) = \frac{\cosh x - x\sinh x - 1}{x^2} $$
Write the function so that it is easier to see the secant being cubed.
$$ \frac d {dx}\left(\sech^3 x\right) = \frac d {dx}\left[(\sech x)^3\right] $$
Differentiate using the chain rule.
$$ \begin{align*} \frac d {dx}\left[(\sech x)^3\right] & = 3(\sech x)^2\cdot \frac d {dx}\left(\sech x\right)\\[6pt] & = 3(\sech x)^2\cdot (-\sech x\tanh x)\\[6pt] & = -3\sech^3 x\tanh x \end{align*} $$
$$\displaystyle \frac d {dx}\left(\sech^3 x\right) = -3\sech^3 x\tanh x$$
Rewrite the function so the powers are more explicit.
$$ f(x) = (\tanh 7x)^3(\sech 7x)^2 $$
Identify the factors in the function.
$$ f(x) = \blue{(\tanh 7x)^3}\red{(\sech 7x)^2} $$
Differentiate using the product rule . Note that each factor requires the chain rule to differentiate.
$$ \begin{align*} f'(x) & = \blue{3(\tanh 7x)^2\cdot \frac d {dx}\left(\tanh 7x\right)}(\sech 7x)^2 + (\tanh 7x)^3\red{\cdot 2(\sech 7x)^1\cdot \frac d {dx}\left(\sech 7x\right)}\\[6pt] & = \blue{3(\tanh 7x)^2\left(7\sech^2 7x\right)}(\sech 7x)^2 + (\tanh 7x)^3\red{\cdot 2(\sech 7x)\left(-7\sech 7x\tanh 7x\right)}\\[6pt] & = 21\tanh^2 7x\sech^4 7x - 14\tanh^4 7x\sech^2 7x \end{align*} $$
Simplify by factoring.
$$ \begin{align*} f'(x) & = 21\blue{\tanh^2 7x}\sech^4 7x - 14\blue{\tanh^4 7x}\sech^2 7x\\[6pt] & = \blue{\tanh^2 7x}\left(21\red{\sech^4 7x} - 14\tanh^2 7x\,\red{\sech^2 7x}\right)\\[6pt] & = \tanh^2 7x\,\red{\sech^2 7x}\left(21\sech^2 7x - 14\tanh^2 7x\right)\\[6pt] & = 7\tanh^2 7x\sech^2 7x\left(3\sech^2 7x - 2\tanh^2 7x\right) \end{align*} $$
$$ \displaystyle f'(x) = 7\tanh^2 7x\sech^2 7x\left(3\sech^2 7x - 2\tanh^2 7x\right) $$
Rewrite the function so the square-root is in exponent notation.
$$ \frac d {dx}\left(\sqrt{1 - \sinh 3x}\right) = \frac d {dx}\left((1 - \sinh 3x)^{1/2}\right) $$
Differentiate using the chain rule. Simplify your result.
$$ \begin{align*} \frac d {dx}\left((1 - \sinh 3x)^{1/2}\right) & = \frac 1 2(1-\sinh 3x)^{-1/2}\cdot (-3\cosh 3x)\\[6pt] & = -\frac 3 2(1-\sinh 3x)^{-1/2}\cosh 3x\\[6pt] & = -\frac 3 2\cdot \frac 1 {\sqrt{1-\sinh 3x}}\cdot \cosh 3x\\[6pt] & = -\frac{3\cosh 3x}{2\sqrt{1-\sinh 3x}} \end{align*} $$
$$ \displaystyle \frac d {dx}\left(\sqrt{1 - \sinh 3x}\right)= -\frac{3\cosh 3x}{2\sqrt{1-\sinh 3x}} $$
Differentiate using the chain rule.
$$ \begin{align*} f'(x) & = e^{\tanh x}\cdot \frac d {dx}\left(\tanh x\right)\\[6pt] & = e^{\tanh x}\cdot \sech^2 x \end{align*} $$
$$f'(x) = e^{\tanh x}\cdot \sech^2 x$$
