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Find the derivative using logarithmic differentiation method $\frac{d}{dx}\left(\tan\left(x+1\right)\right)$

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Solution

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Final Answer

$\sec\left(x+1\right)\sec\left(x+1\right)$
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Step-by-step Solution

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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 $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{1}{\tan\left(x+1\right)}\frac{d}{dx}\left(\tan\left(x+1\right)\right)$
6

The derivative of the linear function is equal to $1$

$\frac{y^{\prime}}{y}=\frac{1}{\tan\left(x+1\right)}\frac{d}{dx}\left(\tan\left(x+1\right)\right)$
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

$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(x\right)\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
10

The derivative of the linear function is equal to $1$

$\frac{y^{\prime}}{y}=\frac{1}{\tan\left(x+1\right)}\sec\left(x+1\right)^2$
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

$\sec\left(x+1\right)\csc\left(x+1\right)\tan\left(x+1\right)$
16

Simplify the derivative

$\sec\left(x+1\right)\sec\left(x+1\right)$

Final Answer

$\sec\left(x+1\right)\sec\left(x+1\right)$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Find the derivativeFind derivative of tan(x+1) using the product ruleFind derivative of tan(x+1) using the quotient ruleFind derivative of tan(x+1) using the definition

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Function Plot

Plotting: $\sec\left(x+1\right)\sec\left(x+1\right)$

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0
a
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d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
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×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Inequalities

In mathematics, an inequation is a statement that an inequality holds between two values. It is usually written in the form of a pair of expressions denoting the values in question, with a relational sign between them indicating the specific inequality relation.

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