Final Answer
$\sec\left(x+1\right)\csc\left(x+1\right)\tan\left(x+1\right)$
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Step-by-step Solution
$\frac{d}{dx}\left(\tan\left(x+1\right)\right)$
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Find the derivative Find the derivative using the product rule Find the derivative using the quotient rule Logarithmic Differentiation Find the derivative using the definition Suggest another method or feature
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1
To derive the function $\tan\left(x+1\right)$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\tan\left(x+1\right)$
2
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\tan\left(x+1\right)\right)$
3
Apply logarithm properties to both sides of the equality
$\ln\left(y\right)=\ln\left(\tan\left(x+1\right)\right)$
4
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(\ln\left(\tan\left(x+1\right)\right)\right)$
5
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{1}{\tan\left(x+1\right)}\frac{d}{dx}\left(\tan\left(x+1\right)\right)$
Intermediate steps
The derivative of the linear function is equal to $1$
$1\left(\frac{1}{y}\right)$
Any expression multiplied by $1$ is equal to itself
$\frac{1}{y}$
6
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{\tan\left(x+1\right)}\frac{d}{dx}\left(\tan\left(x+1\right)\right)$
Explain more
7
The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if ${f(x) = tan(x)}$, then ${f'(x) = sec^2(x)\cdot D_x(x)}$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(x+1\right)\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
8
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\left(\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)\right)\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
9
The derivative of the constant function ($1$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(x\right)\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
Intermediate steps
The derivative of the linear function is equal to $1$
$1\left(\frac{1}{y}\right)$
Any expression multiplied by $1$ is equal to itself
$\frac{1}{y}$
The derivative of the linear function is equal to $1$
$1\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
Any expression multiplied by $1$ is equal to itself
$\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
10
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
Explain more
11
Multiply the fraction and term
$\frac{y^{\prime}}{y}=\frac{\sec\left(x+1\right)^2}{\tan\left(x+1\right)}$
12
Simplify $\frac{\sec\left(x+1\right)^2}{\tan\left(x+1\right)}$ by applying trigonometric identities
$\frac{y^{\prime}}{y}=\sec\left(x+1\right)\csc\left(x+1\right)$
13
Multiply both sides of the equation by $y$
$y^{\prime}=y\sec\left(x+1\right)\csc\left(x+1\right)$
14
Substitute $y$ for the original function: $\tan\left(x+1\right)$
$y^{\prime}=\sec\left(x+1\right)\csc\left(x+1\right)\tan\left(x+1\right)$
15
The derivative of the function results in
$\sec\left(x+1\right)\csc\left(x+1\right)\tan\left(x+1\right)$
Final Answer$\sec\left(x+1\right)\csc\left(x+1\right)\tan\left(x+1\right)$