1
Here, we show you a step-by-step solved example of advanced differentiation. This solution was automatically generated by our smart calculator:
$\frac{d}{dx}\left(cosh\:x\right)^{arccosh\:x}$
2
To derive the function $\mathrm{cosh}\left(x\right)^{\mathrm{arccosh}\left(x\right)}$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\mathrm{cosh}\left(x\right)^{\mathrm{arccosh}\left(x\right)}$
3
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\mathrm{cosh}\left(x\right)^{\mathrm{arccosh}\left(x\right)}\right)$
4
Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$
$\ln\left(y\right)=\mathrm{arccosh}\left(x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)$
5
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)\right)$
6
Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\mathrm{arccosh}\left(x\right)$ and $g=\ln\left(\mathrm{cosh}\left(x\right)\right)$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{d}{dx}\left(\ln\left(\mathrm{cosh}\left(x\right)\right)\right)\mathrm{arccosh}\left(x\right)$
Passi intermedi
The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1}{\mathrm{cosh}\left(x\right)}\frac{d}{dx}\left(\mathrm{cosh}\left(x\right)\right)\mathrm{arccosh}\left(x\right)$
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The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1}{\mathrm{cosh}\left(x\right)}\frac{d}{dx}\left(\mathrm{cosh}\left(x\right)\right)\mathrm{arccosh}\left(x\right)$
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8
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1}{\mathrm{cosh}\left(x\right)}\frac{d}{dx}\left(\mathrm{cosh}\left(x\right)\right)\mathrm{arccosh}\left(x\right)$
9
Taking the derivative of hyperbolic cosine
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1}{\mathrm{cosh}\left(x\right)}\frac{d}{dx}\left(x\right)\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)$
10
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1}{\mathrm{cosh}\left(x\right)}\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)$
Passi intermedi
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}$
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Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}$
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12
Apply the trigonometric identity: $\frac{\mathrm{sinh}\left(\theta \right)}{\mathrm{cosh}\left(\theta \right)}$$=\mathrm{tanh}\left(\theta \right)$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
13
Apply the formula: $\mathrm{arccosh}\left(\theta \right)$$=\ln\left(\theta +\sqrt{\theta ^2-1}\right)$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(x+\sqrt{x^2-1}\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
Passi intermedi
The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1}{\mathrm{cosh}\left(x\right)}\frac{d}{dx}\left(\mathrm{cosh}\left(x\right)\right)\mathrm{arccosh}\left(x\right)$
14
The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\frac{d}{dx}\left(x+\sqrt{x^2-1}\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{d}{dx}\left(\sqrt{x^2-1}\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{d}{dx}\left(\sqrt{x^2-1}\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}\left(x^2-1\right)^{\frac{1}{2}-1}\frac{d}{dx}\left(x^2-1\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
Simplify the addition $\frac{1}{2}-1$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}\left(x^2-1\right)^{\frac{1-2}{2}}\frac{d}{dx}\left(x^2-1\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
Add the values $1$ and $-2$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}\frac{d}{dx}\left(x^2-1\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
16
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}\frac{d}{dx}\left(x^2-1\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
The derivative of the constant function ($-1$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}\frac{d}{dx}\left(x^2\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}\frac{d}{dx}\left(x^2\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}x^{\left(2-1\right)}$
Subtract the values $2$ and $-1$
$2\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}x$
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The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+2\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
Multiply the fraction and term in $2\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}x$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{2\cdot 1}{2}\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{2}{2}\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+1\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Multiply the fraction and term in $2\frac{1}{2}\left(x^2-1\right)^{-\frac{1}{2}}x$
$\frac{y^{\prime}}{y}=\frac{1}{x+\sqrt{x^2-1}}\left(1+\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{1\left(1+\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{\left(1+\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{\left(1+\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
$\frac{1}{\left(x^2-1\right)^{\left|-\frac{1}{2}\right|}}x$
Multiplying the fraction by $x$
$\frac{x}{\left(x^2-1\right)^{\left|-\frac{1}{2}\right|}}$
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Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
$\frac{y^{\prime}}{y}=\frac{\left(1+\frac{1}{\sqrt{x^2-1}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Passi intermedi
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{\left(1+\frac{1x}{\sqrt{x^2-1}}\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{1\left(1+\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{1\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\mathrm{arccosh}\left(x\right)\right)\ln\left(\mathrm{cosh}\left(x\right)\right)+\frac{\mathrm{arccosh}\left(x\right)\mathrm{sinh}\left(x\right)}{\mathrm{cosh}\left(x\right)}$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{\left(1+\left(x^2-1\right)^{-\frac{1}{2}}x\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{\left(1+\frac{x}{\sqrt{x^2-1}}\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{\left(1+\frac{x}{\sqrt{x^2-1}}\right)\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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23
Combine all terms into a single fraction with $\sqrt{x^2-1}$ as common denominator
$\frac{y^{\prime}}{y}=\frac{\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
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Simplify the fraction $\frac{\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\ln\left(\mathrm{cosh}\left(x\right)\right)}{x+\sqrt{x^2-1}}$ by $x+\sqrt{x^2-1}$
$\frac{y^{\prime}}{y}=\frac{\ln\left(\mathrm{cosh}\left(x\right)\right)}{\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)$
25
Multiply both sides of the equation by $y$
$y^{\prime}=\left(\frac{\ln\left(\mathrm{cosh}\left(x\right)\right)}{\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)\right)y$
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Substitute $y$ for the original function: $\mathrm{cosh}\left(x\right)^{\mathrm{arccosh}\left(x\right)}$
$y^{\prime}=\left(\frac{\ln\left(\mathrm{cosh}\left(x\right)\right)}{\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)\right)\mathrm{cosh}\left(x\right)^{\mathrm{arccosh}\left(x\right)}$
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The derivative of the function results in
$\left(\frac{\ln\left(\mathrm{cosh}\left(x\right)\right)}{\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)\right)\mathrm{cosh}\left(x\right)^{\mathrm{arccosh}\left(x\right)}$
Risposta finale al problema
$\left(\frac{\ln\left(\mathrm{cosh}\left(x\right)\right)}{\sqrt{x^2-1}}+\mathrm{arccosh}\left(x\right)\mathrm{tanh}\left(x\right)\right)\mathrm{cosh}\left(x\right)^{\mathrm{arccosh}\left(x\right)}$