1
Here, we show you a step-by-step solved example of derivatives of hyperbolic trigonometric functions. This solution was automatically generated by our smart calculator:
$\frac{d}{dx}\left(csch^2\left(4x^3+1\right)\right)$
Passi intermedi
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2\mathrm{csch}\left(4x^3+1\right)^{2-1}\frac{d}{dx}\left(\mathrm{csch}\left(4x^3+1\right)\right)$
Add the values $2$ and $-1$
$2\mathrm{csch}\left(4x^3+1\right)^{1}\frac{d}{dx}\left(\mathrm{csch}\left(4x^3+1\right)\right)$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2\mathrm{csch}\left(4x^3+1\right)^{2-1}\frac{d}{dx}\left(\mathrm{csch}\left(4x^3+1\right)\right)$
Subtract the values $2$ and $-1$
$2\mathrm{csch}\left(4x^3+1\right)^{1}\frac{d}{dx}\left(\mathrm{csch}\left(4x^3+1\right)\right)$
2
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2\mathrm{csch}\left(4x^3+1\right)^{1}\frac{d}{dx}\left(\mathrm{csch}\left(4x^3+1\right)\right)$
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3
Any expression to the power of $1$ is equal to that same expression
$2\frac{d}{dx}\left(\mathrm{csch}\left(4x^3+1\right)\right)\mathrm{csch}\left(4x^3+1\right)$
4
Taking the derivative of hyperbolic cosecant
$-2\frac{d}{dx}\left(4x^3+1\right)\mathrm{csch}\left(4x^3+1\right)\mathrm{csch}\left(4x^3+1\right)\mathrm{coth}\left(4x^3+1\right)$
5
When multiplying two powers that have the same base ($\mathrm{csch}\left(4x^3+1\right)$), you can add the exponents
$-2\mathrm{csch}\left(4x^3+1\right)^2\frac{d}{dx}\left(4x^3+1\right)\mathrm{coth}\left(4x^3+1\right)$
Passi intermedi
The derivative of the constant function ($1$) is equal to zero
$-2\mathrm{csch}\left(4x^3+1\right)^2\frac{d}{dx}\left(4x^3\right)\mathrm{coth}\left(4x^3+1\right)$
6
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$-2\mathrm{csch}\left(4x^3+1\right)^2\frac{d}{dx}\left(4x^3\right)\mathrm{coth}\left(4x^3+1\right)$
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Passi intermedi
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$-2\cdot 4\mathrm{csch}\left(4x^3+1\right)^2\frac{d}{dx}\left(x^3\right)\mathrm{coth}\left(4x^3+1\right)$
$-8\mathrm{csch}\left(4x^3+1\right)^2\frac{d}{dx}\left(x^3\right)\mathrm{coth}\left(4x^3+1\right)$
7
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$-8\mathrm{csch}\left(4x^3+1\right)^2\frac{d}{dx}\left(x^3\right)\mathrm{coth}\left(4x^3+1\right)$
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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 $f'(x) = nx^{n-1}$
$-24\mathrm{csch}\left(4x^3+1\right)^2x^{\left(3-1\right)}\mathrm{coth}\left(4x^3+1\right)$
Subtract the values $3$ and $-1$
$-24\mathrm{csch}\left(4x^3+1\right)^2x^{2}\mathrm{coth}\left(4x^3+1\right)$
8
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$-8\cdot 3\mathrm{csch}\left(4x^3+1\right)^2x^{2}\mathrm{coth}\left(4x^3+1\right)$
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9
Multiply $-8$ times $3$
$-24\mathrm{csch}\left(4x^3+1\right)^2x^{2}\mathrm{coth}\left(4x^3+1\right)$
Risposta finale al problema
$-24\mathrm{csch}\left(4x^3+1\right)^2x^{2}\mathrm{coth}\left(4x^3+1\right)$