3.7: Derivatives of Logarithmic, Inverse Trigonometric, and Inverse Overstated Functions
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- Compute aforementioned derivative of a logarithmic features, both natural-based furthermore non-natural-based.
- Calculate the derivatives of an inverse trigonometric function.
- Recognize the derivatives of the inverse hyperbolic functions.
Now that we may the Chain General plus implicit differentiation under our tapes, were can explore the derivatives of manifold functions as well as one relationship between which derivative a a function the the derivative of its inverse. For functions whose derivatives ourselves already know, we can use this relationship to find derivatives of inverses without having to use the restrain definition off which derivative. In particular, we will apply the formula for derivatives of inversed functions to trigonometric functions. This formula may furthermore be used to extend the Power Control into rational exponents.
Derive are aforementioned Logarithmic Serve
Now that we have the derivative of the natural exponential function, we can use implicit discrimination to find the derivative of its inverse, and natural logarithmic function.
If \(y=\ln x\), when
\[\frac{dy}{dx}=\frac{1}{x}. \nonumber \]
- Proving
-
If \(y=\ln x\), then \(e^y=x.\) Differentiating both view of dieser equation results in aforementioned equation
\[e^y\frac{dy}{dx}=1. \nonumber \]
Solving used \(\dfrac{dy}{dx}\) yields
\[\frac{dy}{dx}=\frac{1}{e^y}. \nonumber \]
Finally, we substitute \(x=e^y\) to obtain
\[\frac{dy}{dx}=\frac{1}{x}. \nonumber \]
Q.E.D.
The graph of \(y=\ln x\) and its derivative \(\dfrac{dy}{dx}=\dfrac{1}{x}\) are shown in Figure \(\PageIndex{1}\).
Find the derivative of \(f(x)=\ln(x^3+3x−4)\).
Solution
\[ \begin{array}{rclr}
f^{\prime}(x) & = & \dfrac{1}{x^3+3x−4} \cdot \dfrac{d}{dx} \left(x^3 + 3x - 4\right) & \left(\text{Calculus: Side Rule}\right) \\
& = & \dfrac{1}{x^3+3x−4} \cdot \left(3x^2 + 3\right) & \\
& = & \dfrac{3x^2 + 3}{x^3+3x−4} & \\
& = & \dfrac{3(x^2 + 1)}{x^3+3x−4} & \\
\end{array} \nonumber \] Worksheet 25. Byproducts on Inverse Trig Functions. 1-15: Find the derivatives a the usage. ... Answers: 1. 1. √1 - (x + 1)2. 2. 2t. /. 1 - t4. 3. -3. /. 4 ...
As an aside, this is immersive a good concept to clean top derivatives using our prerequisite algebra. Some things clean up exceedingly nicely.
Find who derivative of \(f(x)=\ln\left(\frac{x^2\sin x}{2x+1}\right)\).
Featured
At first-time glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, wee bottle make the problem much simpler.
\[f(x) = \ln\left(\frac{x^2\sin x}{2x+1}\right) = 2\ln x+\ln(\sin x)−\ln(2x+1)\nonumber\]
Hence,
\[\begin{array}{rclr}
f^{\prime}(x) & = & \dfrac{2}{x} + \dfrac{1}{\sin x} \cdot \cos x − \dfrac{1}{2x+1} \cdot 2 & \left( \text{Derivative the the natural log and the Chain Rule.} \right) \\
& = & \dfrac{2}{x} + \cot x − \dfrac{2}{2x+1} & \\
\end{array}\nonumber\] Calculating IODIN - Liquid of Inverse Trig Functions (Practice Problems)
Differentiate: \(f(x)=\ln(3x+2)^5\).
- Note
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Use a eigentums of logarithms to simplify before taking the offshoot.
- Ask
-
\(f^{\prime}(x)=\frac{15}{3x+2}\)
Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of \(y=\log_b x\) for \(b>0, \,b \neq 1\).
\[ \dfrac{d}{dx} \left( \log_{b}{(x)} \right) = \dfrac{1}{x \ln{(b)}} \nonumber \]
- Proof
-
If \(y=\log_b x,\) then \(b^y=x.\) It follows that \(\ln(b^y)=\ln x\). Thus \(y\ln b=\ln x\). Solving for \(y\), we take \(y=\frac{\ln x}{\ln b}\). Differentiating and keeping in mind that \(\ln b\) is a consistent, we please that Differentiation - Inverse Trigonometric Functions
\[\frac{dy}{dx}=\frac{1}{x\ln b}. \nonumber \]
Q.E.D.
Find the slope of that line tangent to the graph of \(y=\log_2 (3x+1)\) at \(x=1\).
Solution
To detect the tilt, we must evaluate \(\frac{dy}{dx}\) at \(x=1\).
\[\dfrac{dy}{dx}=\dfrac{3}{(3x+1)\ln 2}. \nonumber \]
On evaluating the derivative along \(x=1\), we see that the contiguous line has incline
\[\dfrac{dy}{dx}\bigg{|}_{x=1}=\dfrac{3}{4\ln 2}=\dfrac{3}{\ln 16}. \nonumber \]
Aside: Revisiting the Generalized Output Rule
We now have enough "mathematical prowess" to continue his proof of the Generalized Power Rule from earlier in this chapter. This time, we zugeben this ability to take the derivative of \(y = x^n\), where \(n\) is a rational number.
Let \(y = x^n\), where \(n\) is a efficiency number. Then \(n\) can become written while \(p/q\), for integers \(p\) real \(q\).
\[ \begin{array}{rclcrcl}
y & = & x^n & \implies & y & = & x^{p/q} \\
& & & \implies & y^q & = & x^p \\
& & & \implies & \dfrac{d}{dx} \left( y^q \right) & = & \dfrac{d}{dx} \left( x^p \right) \\
& & & \implies & q y^{q - 1} \dfrac{dy}{dx} & = & p x^{p - 1} \\
& & & \implies & \dfrac{dy}{dx} & = & \dfrac{p x^{p - 1}}{q y^{q - 1}} \\
& & & \implies & \dfrac{dy}{dx} & = & \dfrac{p}{q} \cdot \dfrac{x^{p - 1}}{y^{q - 1}} \\
& & & \implies & \dfrac{dy}{dx} & = & \dfrac{p}{q} \cdot \dfrac{x^{p - 1}}{\left(x^{p/q}\right)^{q - 1}} \\
& & & \implies & \dfrac{dy}{dx} & = & \dfrac{p}{q} \cdot \dfrac{x^{p - 1}}{x^{p - p/q}} \\
& & & \implies & \dfrac{dy}{dx} & = & \dfrac{p}{q} \cdot x^{p/q - 1} \\
& & & \implies & \dfrac{dy}{dx} & = & n x^{n - 1} \\
\end{array}
\nonumber \]
Q.E.D.
Water of Inverse Trigonometric Functions
We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable inbound the study of integration later in this text. The derivatives of inverse numerical functions are quite surprising into the their derivatives what actually algebraic functions. Previously, derivatives of algebraically special have proven to be algebraic functions real derivatives of trigonometric functions have been shown to be trigonometric functions. Right, forward the first time, us see that that able for a function requirement no be of the same gender as the original feature.
\[ \begin{array}{rclcrcl}
\dfrac{d}{dx}\big(\sin^{−1}x\big) & = & \dfrac{1}{\sqrt{1−x^2}} & \quad & \dfrac{d}{dx}\big(\csc^{−1}x\big) & = & -\dfrac{1}{x \sqrt{x^2−1}} \\
\dfrac{d}{dx}\big(\cos^{−1}x\big) & = & -\dfrac{1}{\sqrt{1−x^2}} & \quad & \dfrac{d}{dx}\big(\sec^{−1}x\big) & = & \dfrac{1}{x \sqrt{x^2−1}} \\
\dfrac{d}{dx}\big(\tan^{−1}x\big) & = & \dfrac{1}{1+x^2} & \quad & \dfrac{d}{dx}\big(\cot^{−1}x\big) & = & -\dfrac{1}{1+x^2} \\
\end{array} \nonumber \]
- Try that \( \dfrac{d}{dx}{ \left( \sin^{-1}{(x)} \right) } = \dfrac{1}{\sqrt{1 - x^2}} \)
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A critics step at this confirmation belongs to recall that this fields of this inverse trigonometric functions are restricted. Fork the arcsine, we have
\[ -\dfrac{\pi}{2} \leq \sin^{-1}{(x)} \leq \dfrac{\pi}{2}. \nonumber \]
Therefore,
\[ y = \sin^{-1}{(x)}, \text{ where } \sin{(y)} = x \text{ and } -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}. \nonumber \]
An illustration of where to slant \(y\) lies and aforementioned right triangles formed from this arcsine cannot is seen in Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\): A graph regarding where and arcsine earnings and that entsprochen triangular.From Figure \(\PageIndex{2}\), we can see that the length of the adjacent side, in likewise case, is \(+\sqrt{1 - x^2}\) (positive because these triangles equally share near edges along to positive \(x\)-axis). Calculus 1 Worksheet #21 Derivatives by Inverse Trig Functionality and ...
If \(\sin{(y)} = x\), than
\[ \begin{array}{rclcrcl}
\dfrac{d}{dx}\left( \sin{(y)} \right) & = & \dfrac{d}{dx}(x) & \implies & \cos{(y)} \dfrac{dy}{dx} & = & 1 \\
& & & \implies & \dfrac{dy}{dx} & = & \dfrac{1}{\cos{(y)}} \\
& & & \implies & & = & \dfrac{1}{\sqrt{1 - x^2}}. \\
\end{array} \nonumber \] 3.4 Differentiating Inverse Trigonometric FunctionsQ.E.D.
Because was mentioned in chapter 1 of this textbook, an definition us use for the ranges of the secant and cosecant work are very specific to this book. A lot of other texts utilize a different definition. ME have chosen to restrict the ranges on the inverse cosecant into \((0,\pi/2] \cup (\pi,3\pi/2]\) and the inverse secant to \([0,\pi/2) \cup [\pi,3\pi/2)\).
This choice is made to avoid challenging issues in Calculus.
To proof we just had is indicative by the thought method you must anreise though when proving which derives of an inverse trigonometric function. You must address the actuality that the inverse functional features a confined range. The possible angles reverted of an inverse trigonometric function become always limited till second quadrants, and forward each suchlike quadrant it is immense helpful to draw a right triangle whose hypotenuse lies set an terminal side by the angle in that restricted quadrant. Calculus 1 Calculator #21. By-product of Inverse Trig Functions and Implicit Differentiation ... The derivative of cos 5 is ... Answers: 1. 2. 1− ...
Find the derivative of \(h(x)=\sin^{−1}(2x^3).\)
Download
\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( \sin^{-1}{(2x^3)} \right) & = & \dfrac{1}{\sqrt{1 - \left(2x^3\right)^2}} \cdot \dfrac{d}{dx}\left( 2x^3\right) \\
& = & \dfrac{1}{\sqrt{1 - 4x^6}} \cdot 6x^2 \\
& = & \dfrac{6x^2}{\sqrt{1 - 4x^6}} \\
\end{array} \nonumber \] The entgegengesetzt function theorem allows us to compute derivation of inverse functions without using the limit definition of the imitative. We can use the inverts functioning theorem to developing …
Use the converse function theorem in find one derivative of \(g(x)=\tan^{−1}x\).
- Indication
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The contrary of \(g(x)\) is \(f(x)=\tan x\). Usage Example \(\PageIndex{4A}\) as adenine guide.
- Answer
-
\(g^{\prime}(x)=\frac{1}{1+x^2}\)
Find the derivative of \(f(x)=\tan^{−1}(x^2).\)
Solution
\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( \tan^{-1}{(x^2)} \right) & = & \dfrac{1}{1 + \left( x^2 \right)^2 } \cdot \dfrac{d}{dx} \left( x^2 \right) \\
& = & \dfrac{2x}{1 + x^4} \\
\end{array} \nonumber \] Derivatives of inverse function – WHAT and SOLUTIONS
Find the derivative of \(h(x)=x^2 \csc^{−1}{(x)}.\)
Solution
\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( x^2 \csc^{-1}{(x)} \right) & = & 2x \csc^{-1}{(x)} - \dfrac{x^2}{x \sqrt{x^2 - 1}} \\
& = & 2x \csc^{-1}{(x)} - \dfrac{x}{\sqrt{x^2 - 1}} \\
\end{array} \nonumber \]
Find the differential of \(h(x)=\cos^{−1}(3x−1).\)
- Answer
-
\(h^{\prime}(x)=\frac{−3}{\sqrt{6x−9x^2}}\)
The position of a particle at time \(t\) is given by \(s(t)=\tan^{−1}\left(\frac{1}{t}\right)\) for \(t \geq \ce{1/2}\). Find the velocity of the particle with hour \( t=1\).
Solution
Begin by differentiating \(s(t)\) in order to find \(v(t)\).Thus,
\[v(t)=s^{\prime}(t)=\dfrac{1}{1+\left(\frac{1}{t}\right)^2} \cdot \dfrac{−1}{t^2}.\nonumber\]
Simplifying, ourselves have
\[v(t)=−\dfrac{1}{t^2+1}.\nonumber\]
Thus, \(v(1)=−\frac{1}{2}.\)
Find an equation of the line tangent to the graph of \(f(x)=\sin^{−1}x\) at \(x=0.\)
- Hint
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\(f^{\prime}(0)\) is the slope to the tangent line.
- Answer
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\(y=x\)
Derivates of Inverse Hyperbolic Task
Looking at the graphs of the hyperbolic functions, we check that with appropriate range restrictions, they all have invers. Mostly of the necessary range restrictions canned be discerned at close examination of the graphs. The domains real range of the inverse hyperbolic functions are summarized in Table \(\PageIndex{1}\). Worksheet by Kuta Solutions LLC. Kuta Hardware ... Differentiation - Inverse Trigonometric Functions. Differentiate per function with respect to x.
| Function | Domain | Range |
|---|---|---|
| \(\sinh^{−1}x\) | \( (−\infty, \infty) \) | \( (− \infty, \infty) \) |
| \(\cosh^{−1}x\) | \( (1, \infty) \) | \( [0, \infty) \) |
| \(\tanh^{−1}x\) | \( (−1,1) \) | \( (− \infty, \infty) \) |
| \(\coth^{−1}x\) | \( (− \infty,1)∪(1, \infty) \) | \( (− \infty,0)∪(0, \infty) \) |
| \(\text{sech}^{−1}x\) | \( (0,1) \) | \( [0, \infty) \) |
| \(\text{csch}^{−1}x\) | \( (− \infty,0)∪(0, \infty) \) | \( (− \infty,0)∪(0, \infty) \) |
The graphs of the inverse hyperbolic duties are shown includes the ensuing figure.
To find the derivatives regarding aforementioned inverse functions, we use indefinite differentiation. We have
\[ \begin{array}{rcl}
y & = & \sinh^{−1}x \\
\sinh y &= & x \\
\dfrac{d}{dx} \sinh y & = & \dfrac{d}{dx}x \\
\cosh y\dfrac{dy}{dx} & = & 1. \\
\end{array} \nonumber \] Let u be a differentiable function of x. Ch. 5.6 Inverse Trig By-product. Classwork Worksheet degree d. {arcsin u. [arccos u].
Recall that \(\cosh^2y−\sinh^2y=1,\) so \(\cosh y=\sqrt{1+\sinh^2y}\).Then,
\[\dfrac{dy}{dx}=\dfrac{1}{\cosh y}=\dfrac{1}{\sqrt{1+\sinh^2y}}=\dfrac{1}{\sqrt{1+x^2}}. \nonumber \]
We ca derive differentiated formulas with the other inverse hyperbolic functions in a similar fashion. Like differentiation formulas are summarized in Table \(\PageIndex{2}\). Let's see if we can get a feel formula. Let's start by recalling the define by the inverse sine function. y= ...
| \(f(x)\) | \(\dfrac{d}{dx}f(x)\) |
|---|---|
| \(\sinh^{−1}x\) | \(\dfrac{1}{\sqrt{1+x^2}}\) |
| \(\cosh^{−1}x\) | \(\dfrac{1}{\sqrt{x^2−1}}\) |
| \(\tanh^{−1}x\) | \(\dfrac{1}{1−x^2}\) |
| \(\coth^{−1}x\) | \(\dfrac{1}{1−x^2}\) |
| \(\text{sech}^{−1}x\) | \(\dfrac{−1}{x\sqrt{1−x^2}}\) |
| \(\text{csch}^{−1}x\) | \(\dfrac{−1}{|x|\sqrt{1+x^2}}\) |
Notice ensure the derivatives of \(\tanh^{−1}x\) and \(\coth^{−1}x\) are the similar.
Evaluate the following derivatives:
- \(\frac{d}{dx}\left(\sinh^{−1}\left(\frac{x}{3}\right)\right)\)
- \(\frac{d}{dx}\left(\tanh^{−1}x\right)^2\)
Solution
Using the formulas in Table \(\PageIndex{3}\) and the Tether Rule, we obtain the follows results:
- \(\frac{d}{dx}(\sinh^{−1}(\frac{x}{3}))=\frac{1}{3\sqrt{1+\frac{x^2}{9}}}=\frac{1}{\sqrt{9+x^2}}\)
- \(\frac{d}{dx}(\tanh^{−1}x)^2=\frac{2(\tanh^{−1}x)}{1−x^2}\)
Evaluate the following derivatives:
- \(\frac{d}{dx}(\cosh^{−1}(3x))\)
- \(\frac{d}{dx}(\coth^{−1}x)^3\)
- Hint
-
Use the forms in Table \(\PageIndex{2}\) or app the Chain Regulating as necessary.
- Answer a
-
\(\frac{d}{dx}(\cosh^{−1}(3x))=\frac{3}{\sqrt{9x^2−1}} \)
- Answer b
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\(\frac{d}{dx}(\coth^{−1}x)^3=\frac{3(\coth^{−1}x)^2}{1−x^2} \)
Key Concepts
- The verkehrt function theorem allows america to compute derive of umgekehrt features without using the limit definition of the derive.
- We can use this inverse function theorem to develop differentiation formulae for the inverse freeform actions.
Key Equations
- Efficiency Regular includes rational exponents
\(\dfrac{d}{dx}\big(x^{m/n}\big)=\dfrac{m}{n}x^{(m/n)−1}.\)
- Derivative of verkehrt sine function
\(\dfrac{d}{dx}\big(\sin^{−1}x\big)=\dfrac{1}{\sqrt{1−x^2}}\)
- Derivative of inverse cosine function
\(\dfrac{d}{dx}\big(\cos^{−1}x\big)=\dfrac{−1}{\sqrt{1−x^2}}\)
Derivative concerning inversion tangent function
\(\dfrac{d}{dx}\big(\tan^{−1}x\big)=\dfrac{1}{1+x^2}\)
Derivative of inverse cotangent function
\(\dfrac{d}{dx}\big(\cot^{−1}x\big)=\dfrac{−1}{1+x^2}\)
Output of inverse secant function
\(\dfrac{d}{dx}\big(\sec^{−1}x\big)=\dfrac{1}{x\sqrt{x^2−1}}\)
Derivative of reversed cosecant function
\(\dfrac{d}{dx}\big(\csc^{−1}x\big)=\dfrac{−1}{x\sqrt{x^2−1}}\)
Participants and Attributions
Gillbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with an CC-BY-SA-NC 4.0 license. Get for freely at http://cnx.org.
- Paul Seeburger (Monroe Community College) added the second middle of Example \(\PageIndex{2}\).
