Exercises

Find the derivative of $y=\tan^{-1}(x^2+1)$

---

Let $u=x^2+1$ then we can apply the Chain Rule:

$$y=\tan^{-1}(u)$$ $$y'=\frac{1}{1+u^2}\frac{du}{dx}$$ $$y'=\frac{1}{1+u^2}\cdot 2x$$ $$y'=\frac{1}{1+(x^2+1)^2}\cdot 2x$$ $$y'=\frac{2x}{1+(x^2+1)^2}$$ $$y'=\frac{2x}{x^4+2x^2+2}$$

Find the derivative of $y=x\sin^{-1}\left(\frac{1}{x}\right)$.

---

First use the Product Rule.

$$y=x\sin^{-1}\left(\frac{1}{x}\right)$$ $$y=x \frac{d}{dx}\sin^{-1}\left(\frac{1}{x}\right) + \frac{d}{dx}x \cdot \sin^{-1}\left(\frac{1}{x}\right)$$ $$y=x \frac{d}{dx}\sin^{-1}\left(\frac{1}{x}\right) + \sin^{-1}\left(\frac{1}{x}\right)$$

Now let $u=\frac{1}{x}$ so $\frac{du}{dx} = -\frac{1}{x^2}$ and apply the Chain Rule.

$$y=x \frac{d}{dx}\sin^{-1} u + \sin^{-1}\left(\frac{1}{x}\right)$$ $$y=x \frac{1}{\sqrt{1-u^2}}\frac{du}{dx} + \sin^{-1}\left(\frac{1}{x}\right)$$ $$y=\frac{x}{\sqrt{1-u^2}}\left(-\frac{1}{x^2}\right) + \sin^{-1}\left(\frac{1}{x}\right)$$ $$y=-\frac{1}{x\sqrt{1-u^2}} + \sin^{-1}\left(\frac{1}{x}\right)$$ $$y=-\frac{1}{x\sqrt{1-(1/x)^2}} + \sin^{-1}\left(\frac{1}{x}\right)$$ $$y=-\frac{1}{x\sqrt{\frac{x^2-1}{x^2}}} + \sin^{-1}\left(\frac{1}{x}\right)$$ $$y=-\frac{1}{\sqrt{x^2-1}} + \sin^{-1}\left(\frac{1}{x}\right)$$

Find the derivative of $y=\arccos(\ln(x))$.

---

Let $u=\ln x$ and $\frac{du}{dx}=\frac{1}{x}$ and apply the Chain Rule.

$$y=\arccos(\ln(x))$$ $$y=\arccos(u)$$ $$y'=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}$$ $$y'=-\frac{1}{\sqrt{1-\ln^2 x}}\frac{1}{x}$$ $$y'=-\frac{1}{x\sqrt{1-\ln^2 x}}$$

Find the derivative of $y=\ln\left(\arccos(x)\right)$.

---

Let $u=\arccos x$ and $\frac{du}{dx}=-\frac{1}{\sqrt{1-x^2}}$ and apply the Chain Rule.

$$y=\ln\left(\arccos(x)\right)$$ $$y'=\frac{1}{\arccos x} \cdot \frac{du}{dx}$$ $$y'=\frac{1}{\arccos x} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$$ $$y'=-\frac{1}{\sqrt{1-x^2}\arccos x}$$

Find the derivative of $y=\arcsin\left(\frac{x+1}{x-1}\right)$.

---

Let $u=\frac{x+1}{x-1}$ and

$$\frac{du}{dx} = \frac{(x-1) \cdot 1 - (x+1) \cdot 1}{(x-1)^2} = -\frac{2}{(x-1)^2} $$

Then the Chain Rule gives us

$$y = \arcsin\left(\frac{x+1}{x-1}\right)$$ $$y = \arcsin u$$ $$y' = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$$ $$y' = \frac{1}{\sqrt{1-u^2}} \cdot \left(-\frac{2}{(x-1)^2}\right)$$ $$y' = \frac{1}{\sqrt{1-\left(\frac{x+1}{x-1}\right)^2}} \cdot \left(-\frac{2}{(x-1)^2}\right)$$ $$y' = \frac{1}{\sqrt{\left(\frac{(x-1)^2-(x+1)^2}{x-1}\right)^2}} \cdot \left(-\frac{2}{(x-1)^2}\right)$$ $$y' = \frac{1}{\sqrt{\left(\frac{-4x}{x-1}\right)^2}} \cdot \left(-\frac{2}{(x-1)^2}\right)$$ $$y' = \frac{1}{\frac{4x}{x-1}} \cdot \left(-\frac{2}{(x-1)^2}\right)$$ $$y' = \frac{x-1}{4x} \cdot \left(-\frac{2}{(x-1)^2}\right)$$ $$y' = -\frac{1}{2x(x-1)}$$ $$y' = -\frac{1}{2x^2-2x}$$

Find the derivative of $y=\cos^{-1}(\cos(x) + 1)$.

---

Let $u=\cos(x) + 1$ and $\frac{du}{dx} = -\ sin x$. Then the Chain Rule gives us

$$y=\cos^{-1}(\cos(x) + 1)$$ $$y=\cos^{-1} u$$ $$y' = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$ $$y' = -\frac{1}{\sqrt{1-(\cos(x) + 1)^2}} \cdot (-\sin x)$$ $$y' = \frac{1}{\sqrt{1-\cos^2(x) - 2\cos x - 1}} \cdot (\sin x)$$ $$y' = \frac{\sin x}{\sqrt{-\cos^2(x) - 2\cos x}}$$

Find $y'$ for the relation $\tan^{-1}(xy)=xy$.

$$\tan^{-1}(xy)=xy$$ $$\frac{1}{1+(xy)^2} \frac{d}{dx}(xy) = xy' + 1y$$ $$\frac{1}{1+(xy)^2} (xy' + y) = xy' + y$$ $$\frac{xy' + y}{1+(xy)^2} = xy' + y$$ $$\frac{xy'}{1+(xy)^2} + \frac{y}{1+(xy)^2} = xy' + y$$ $$\frac{xy'}{1+(xy)^2} - xy' = y - \frac{y}{1+(xy)^2}$$ $$y'\left(\frac{x}{1+(xy)^2} - x\right) = \frac{y(1+(xy)^2)}{1+(xy)^2} - \frac{y}{1+(xy)^2}$$ $$y'\frac{x-x(1+(xy)^2)}{1+(xy)^2} = \frac{y+y(xy)^2-y}{1+(xy)^2}$$ $$y'\frac{x(xy)^2)}{1+(xy)^2} = \frac{y(xy)^2}{1+(xy)^2}$$ $$y'(x^3y^2)=x^2y^3$$ $$y'=\frac{y}{x}$$

Find $y'$ for the relation $\sec^{-1}\left(\frac{x}{y}\right)=1$.

$$\sec^{-1}\left(\frac{x}{y}\right)=1$$ $$\frac{1}{\frac{x}{y}\sqrt{\left(\frac{x}{y}\right)^2-1}}\frac{d}{dx}\left(\frac{x}{y}\right)=0$$ $$\frac{1}{\frac{x}{y}\sqrt{\left(\frac{x}{y}\right)^2-1}}\frac{xy'-1\cdot y}{y^2}=0$$ $$\frac{xy'-1\cdot y}{xy\sqrt{\left(\frac{x}{y}\right)^2-1}}=0$$ $$xy'-y=0$$ $$y' = \frac{y}{x}$$

We can be a little casual sometimes about elminating things when we're solving these equations. For example, when I eliminated $xy\sqrt{\left(\frac{x}{y}\right)^2-1}$ from the denominator. It's okay to do that as long as that expression is never equal to 0. If there are places where it is equal to 0 then the derivative will actually be undefined there. A key point to keep in mind here is that you can't pick any values that you like for $x$ and $y$ here. THe only points we care about are ones that satisfy the original equation. To see that it was okay to do what I did, look at the equation:

$$xy\sqrt{\left(\frac{x}{y}\right)^2-1} = 0$$

That equation has three possible solutions:

1. $x = 0$ which isn't possible because $\arcsin(0/y) = \arcsin(0)$ which is undefined for all values of $y$.
2. $y=0$ which also isn't possible because $arcsin(x/0)$ is undefined for all values of $x$.
3. $x/y-1=0$ which only occurs when $x=y$. Whenever $x=y$, the original equation becomes $\arcsin(1) = 1$ which is false.

Since there were no values of $x$ and $y$ that made the expression I eliminated equal to 0, it was okay to multiply both sides by it.

Find $y'$ for the relation $\left(\frac{\tan^{-1} x}{\tan^{-1}y}\right)=x$.

$$\left(\frac{\tan^{-1} x}{\tan^{-1}y}\right)=x$$ $$\frac{\tan^{-1} y \cdot \frac{d}{dx}\tan^{-1} x - \tan^{-1} x \cdot \frac{d}{dx}\tan^{-1} y}{(\tan^{-1} y)^2} = 1$$ $$\frac{\tan^{-1} y \cdot \frac{1}{1+x^2} - \tan^{-1} x \cdot \frac{1}{1+y^2} \cdot y'}{(\tan^{-1} y)^2} = 1$$ $$\frac{\frac{\tan^{-1} y}{1+x^2} - y'\frac{\tan^{-1} x}{1+y^2}}{(\tan^{-1} y)^2} = 1$$ $$\frac{\tan^{-1} y}{1+x^2} - y'\frac{\tan^{-1} x}{1+y^2} = (\tan^{-1} y)^2$$ $$y'\frac{\tan^{-1} x}{1+y^2} = -(\tan^{-1} y)^2 + \frac{\tan^{-1} y}{1+x^2}$$ $$y'\frac{\tan^{-1} x}{1+y^2} = \frac{\tan^{-1} y-(\tan^{-1} y)^2(1+x^2)}{1+x^2}$$ $$y' = \frac{(1+y^2)(\tan^{-1} y-(\tan^{-1} y)^2(1+x^2))}{(1+x^2)\tan^{-1} x}$$

You could arguably simlify this further by multiplying out the numerator but I don't think there's a lot to be gained by it.

Find the derivative of the inverse of $f(x) = x\cos x$ at $x = 0 = f(\pi/2)$.

---

Start by using the formula for the derivative of the inverse of a function:

$$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$$

Then substitute 0 for $x$. Remember that if the point $(x, y)$ satisfies the original function then the point $(y, x)$ satisfies the inverse.

Find the derivative of the inverse of $f(x) = x\cos x$ at $x = 0 = f(\pi/2)$.

---

$$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$$ $$\frac{d}{dx}f^{-1}(0) = \frac{1}{f'(f^{-1}(0))}$$ $$\frac{d}{dx}f^{-1}(0) = \frac{1}{f'(\pi/2)}$$

Since $f'(x) = \cos x - x\sin x$ we have $f'(\pi/2) = \cos (\pi/2) - \frac{\pi}{2} \cdot \sin (\pi/2)=- \frac{\pi}{2}$. This leaves us with

$$\frac{d}{dx}f^{-1}(0) = \frac{1}{-\frac{\pi}{2}} = -\frac{2}{\pi}$$

Find the derivative of the inverse of $f(x) = x^2 \ln x$ at $x = e^2 = f(e)$.

---

$$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$$ $$\frac{d}{dx}f^{-1}(e^2) = \frac{1}{f'(f^{-1}(e^2))}$$ $$\frac{d}{dx}f^{-1}(e^2) = \frac{1}{f'(e)}$$

Since $f'(x) = x + 2x\ln x$ we have $f'(e) = e + 2e\ln e = 3e$. This leaves us with

$$\frac{d}{dx}f^{-1}(e^2) = \frac{1}{3e}$$

Using a graphing program, e.g. desmos.com, to graph the relation $\tan^{-1}(xy)=xy$. How does this compare to the derivative you found for this relation in a question earlier on this page? (You can try graphing the function by hand but it's a fairly unusual result.)

---

First, recall that, for this relation, $y'=\frac{y}{x}$. Your graph should be just two lines - the x- and y-axes.

Points on the y-axis have the form $(0,y)$. For those values, our formula for $y'$ says that the slope of the tangent line is undefined which is what we would expect for a vertical line like the y-axis.

Points on the x-axis, on the other hand, have the form $(x, 0)$. For those values, our formula says that $y' = \frac{0}{x}$ which is equal to 0 which is consistent with what we would expect for a horizontal line like the x-axis.

The origin, $(0, 0)$ is an exception to the previous paragraph but, at that point, we would have $y' = \frac{0}{0}$ which is going to be undefined which is also what we would expect at a point where the graph makes a sharp angle.

Derive the formula for the derivative of $f(x) = \csc^{-1} x$

Let $g(x) = \csc x$ so $f(x) = g^{-1}(x)$. Then we have

$$\frac{d}{dx}g^{-1}(x) = \frac{1}{g'(g^{-1}(x))}$$ $$\frac{d}{dx}\csc^{-1}(x) = \frac{1}{-\csc(g^{-1}(x))\cot(g^{-1}(x))}$$ $$\frac{d}{dx}\csc^{-1}(x) = -\frac{1}{\csc(\csc^{-1} x)\cot(\csc^{-1} x)}$$ $$\frac{d}{dx}\csc^{-1}(x) = -\frac{1}{x\cot(\csc^{-1} x)}$$

To simplify the $\cot(\csc^{-1} x)$ part, look at the following diagram. It was drawn so that the cosecant of the angle $\theta$ is $x$, i.e. $\csc^{-1} x = \theta$. The bottom side was filled in using the Pythagorean Theorem.

If we substitute that into our derivative equation, it becomes

$$\frac{d}{dx}\csc^{-1}(x) = -\frac{1}{x\cot(\theta)}$$ $$\frac{d}{dx}\csc^{-1}(x) = -\frac{1}{x\sqrt{x^2-1}}$$

Derive the formula for the derivative of $y = \cot^{-1} x$ by differentiating both sides of $\cot y = x$.

---

$$\cot y = x$$ $$-\csc^2 y y' = 1$$ $$y' = -\frac{1}{\csc^2 y }$$

Going back to the original equation $y = \cot^{-1} x$, $y$ is an angle whose cotangent is $x$. This relationship is illustrated in the triangle below with the hypotenuse filled in using the Pythagorean Theorem.

This means that $\csc y = \sqrt{x^2+1}$ which makes our derivative equation

$$y' = -\frac{1}{(\sqrt{x^2+1})^2 }$$ $$y' = -\frac{1}{x^2+1}$$

Use the trigonometric identity $\sec^{-1}x= \frac{\pi}{2} - \csc^{-1} x$ to derive a formula for the derivative of $\sec^{-1}x$ using the derivative of $\csc^{-1}x$.

---

$$\sec^{-1}x= \frac{\pi}{2} - \csc^{-1} x$$ $$\frac{d}{dx}\sec^{-1}x= 0 - \left(-\frac{1}{x\sqrt{x^2-1}}\right)$$ $$\frac{d}{dx}\sec^{-1}x= \frac{1}{x\sqrt{x^2-1}}$$

Simplify $f(x) = \cos(\sin^{-1} x)$ trigonometrically then find the derivatives of both versions of the functions and confirm that both methods give the same result.

---

Applying the definition for the derivative of the inverse cosine gives us

$$f(x) = \cos(\sin^{-1} x)$$ $$f'(x) = -\sin(\sin^{-1} x)\frac{d}{dx}\sin^{-1} x$$ $$f'(x) = -x \cdot \frac{1}{\sqrt{1-x^2}}$$ $$f'(x) = -\frac{x}{\sqrt{1-x^2}}$$

On the other hand, if let $\theta = \sin^{-1}x$ then $\sin \theta = x$ and the relationship is illustrated in the triangle below with the length of the bottom side filled in using the Pythagorean Theorem.

This makes our equation

$$f(x) = \cos(\sin^{-1} x) = \cos heta = \sqrt{1-x^2}$$

so

$$\begin{aligned} f(x) &= (1-x^2)^{1/2} \\ &= \frac{1}{2}\cdot (1-x^2)^{-1/2} \frac{d}{dx}(1-x^2) \\ &= \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) \\ &= -\frac{x}{\sqrt{1-x^2}} \end{aligned}$$