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 derivatives product rule quotient rule chain rule implicit differentiation

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

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There are times when we need to know the derivative of an inverse function but it is not possible to calculate the actual inverse function.
For a full lecture on this topic, we recommend this video.

### Prof Leonard - Calculus 2 Lecture 6.2: Derivatives of Inverse Functions [44mins-8secs]

video by Prof Leonard

When we are not able to calculate the inverse of a function but we need to know the derivative of the inverse at a point, we can use implicit differentiation to come up with a formula to do the calculation.

We need to be careful though to make sure that the function actually has an inverse. In order for a function to have an inverse, it must be one-to-one. One way to make sure it is one-to-one is to check that the function is always either increasing or decreasing in the interval we are concerned about. Of course, using calculus, we can take the derivative and find the intervals where the function is increasing and decreasing. Let's look at an example.

Example 1

built with GeoGebra

Does the function $$f(x) = x^3 + 7x + 2$$ have an inverse?
Let's take the derivative and see if we can determine anything from it.
$$f'(x) = 3x^2 + 7$$
Notice that the derivative is always positive for all x since $$3x^2$$ is positive and adding a positive 7 keeps the result positive. So, since the function is always increasing, it is one-to-one on it's entire domain and therefore has an inverse. To help you visualise this, here is the graph.

So, what do we do when have a function, we know it has an inverse but we can't find the inverse in order to take the derivative? Well, we can use implicit differentiation to find an expression for the derivative at a point as follows.

$$\begin{array}{rcl} y & = & f(x) \\ f^{-1}(y) & = & f^{-1}(f(x)) \\ f^{-1}(y) & = & x \\ \displaystyle{ \frac{d}{dx} } [ f^{-1}(y) ] & = & \displaystyle{ \frac{d}{dx} } [x] \\ (f^{-1}(y))' \displaystyle{ \frac{dy}{dx} } & = & 1 \\ (f^{-1}(y))' & = & \displaystyle{ \frac{1}{dy/dx} } \\ (f^{-1}(y))' & = & \displaystyle{ \frac{1}{f'(x)} } \\ \end{array}$$

Now we have an expression for the derivative of an inverse function at the point $$(x_1,y_1)$$ on $$f(x)$$ which is the point $$(y_1,x_1)$$ on $$f^{-1}(x)$$. $(f^{-1}(y_1))' = \frac{1}{f'(x_1)}$ The tick marks used to write the derivatives above hide important details of this equation. To more clearly show what is going on, we can write the equation like this. $\frac{d}{dy} [f^{-1}(y)] = \frac{1}{dy/dx}$ Notice that the derivative on the left is with respect to y and the derivative on the right is with respect to x.

This may still be a bit confusing, so let's look at it another way. If we have a function $$f(x)$$ with an inverse $$g(x)$$, then $g'(x) = \frac{1}{f'(g(x))}$ at the point $$(x,g(x))$$.

So how do we use all these equations to solve a problem. Let's go through an example to demonstrate how to keep everything straight.

Example 2

For the function $$f(x)=x^2; x \gt 0$$, calculate $$(f^{-1}(9))'$$.
For this example, it is possible to find the inverse for $$x \gt 0$$ and we will do that later to check our answer. But let's use the above equations to get an answer first. We will build a table to keep track of everything. The first row shows information about $$f(x)$$ and the second row shows information about the inverse of $$f(x)$$.

 $$f(x) = x^2$$ $$f'(x) = 2x$$ $$f'(g(9))$$ $$f^{-1}(x) = g(x)$$ $$(x, g(x) ) = (9,3)$$ $$g'(9)$$

We have a partially filled in table above with some basic information. We are given $$f(x)$$ and the point $$x=9$$ on the inverse function. We changed the notation a bit and wrote $$f^{-1}(x) = g(x)$$. Using $$g(x)$$ makes the notation simpler and easier to use. In the above table we took the derivative of $$f(x)$$ and calculated the point $$(9,3)$$. To get the 3, we note that we are given $$x=9$$ on $$g(x)$$ and we know that since $$f(x)$$ and $$g(x)$$ are inverses, then $$f(x) = 9$$. This means that $$x^2=9 \to x=\pm 3$$ and since the problem stated that $$x \gt 0$$, we know that $$x=3$$. So that gives us the point $$(9,3)$$.

Now we can calculate $$f'(g(x))$$. From the table, $$g(9)=3$$, so $$f'(g(9)) = f'(3)$$ giving us $$f'(g(9)) = 6$$. So we will fill in that spot in the table and the result in the last row and last column, which comes from $$g'(9) = 1/f'(g(9))$$.

 $$f(x) = x^2$$ $$f'(x) = 2x$$ $$f'(g(9)) = 6$$ $$g(x)$$ $$(x, g(x) ) = (9,3)$$ $$g'(9) = 1/6$$

Let's check our answer by directly calculating the derivative of the inverse function.
$$\begin{array}{rcl} g(x) & = & \sqrt{x} \\ g'(x) & = & \displaystyle{ \frac{1}{2x^{1/2}} } \\ g'(9) & = & \displaystyle{ \frac{1}{(2)9^{1/2}} = \frac{1}{6} } \text{✔} \end{array}$$

$$(f^{-1}(9))' = g'(9) = 1/6$$

The key to working these problems correctly is to keep track of the x and y values and knowing what to use when. We think that the using a table, like we did in the second example above, helps.

Before you try some on your own, watch this quick video clip with a simplified explanation.

### Calculus 1-The Derivative of an Inverse Function [mins-secs]

Okay, time for some practice problems.

### Practice

If $$f(x)=x^3+e^x$$ and $$f^{-1}(1)=0$$, calculate the derivative of $$f^{-1}$$ at $$x=1$$. Make sure you check that this function has an inverse.

Problem Statement

If $$f(x)=x^3+e^x$$ and $$f^{-1}(1)=0$$, calculate the derivative of $$f^{-1}$$ at $$x=1$$. Make sure you check that this function has an inverse.

Solution

### 2349 solution video

video by PatrickJMT

If $$f(x) = \cos x + 3x$$ and $$f^{-1}(3\pi/2) = \pi/2$$, calculate the derivative of $$f^{-1}$$ at $$x=3\pi/2$$. Make sure to check that $$f(x)$$ has an inverse.

Problem Statement

If $$f(x) = \cos x + 3x$$ and $$f^{-1}(3\pi/2) = \pi/2$$, calculate the derivative of $$f^{-1}$$ at $$x=3\pi/2$$. Make sure to check that $$f(x)$$ has an inverse.

Solution

### 2350 solution video

video by PatrickJMT

For $$f(x)=2x^3+3x^2+7x+4$$, calculate $$(f^{-1})'(4)$$.

Problem Statement

For $$f(x)=2x^3+3x^2+7x+4$$, calculate $$(f^{-1})'(4)$$.

Solution

### 2351 solution video

For $$f(x)=x^3+2x-1$$, calculate $$(f^{-1})'(2)$$.

Problem Statement

For $$f(x)=x^3+2x-1$$, calculate $$(f^{-1})'(2)$$.

Solution

### 2352 solution video

For $$f(x)=(1/27)(x^5+2x^3)$$, calculate $$(f^{-1})'(-11)$$.

Problem Statement

For $$f(x)=(1/27)(x^5+2x^3)$$, calculate $$(f^{-1})'(-11)$$.

Solution

### 2353 solution video

For $$f(x) = x^3-4/x$$, calculate $$(f^{-1})'(6)$$.

Problem Statement

For $$f(x) = x^3-4/x$$, calculate $$(f^{-1})'(6)$$.

Solution