Tag: Chain Rule

Prove a property of the derivative if arctangent and the logarithm obey a given relation

by RoRi

November 29, 2015

$\arctan \frac{y}{x} = \log \sqrt{x^2 + y^2}$

prove that

$\frac{dy}{dx} = \frac{x+y}{x-y}.$

Proof. First, we consider the derivatives of the left and right side of the given equation. (Treating $y$ as a function of $x$ and remembering to use the chain rule.) So, for the derivative on the left, we have

$\begin{align*} D \left( \arctan \frac{y}{x} \right) &= \left( \frac{-y}{x^2} + \frac{1}{x} \frac{dy}{dx} \right) \left( \frac{1}{1+\left( \frac{y}{x} \right)^2} \right) \\ &= \frac{x \frac{dy}{dx} - y}{x^2+y^2}. \end{align*}$

On the right we have,

$\begin{align*} D \left( \log \sqrt{x^2+y^2} \right) &= \left( x + y \frac{dy}{dx} \right) \left( \frac{1}{x^2+y^2} \right)\\ &= \frac{x + y \frac{dy}{dx}}{x^2+y^2}. \end{align*}$

Now, using the given equation we have

$\begin{align*} \arctan \frac{y}{x} = \log \sqrt{x^2+y^2} && \implies && D \left( \arctan \frac{y}{x} \right) &= D \left( \log \sqrt{x^2+y^2} \right) \\[9pt] && \implies && \frac{x \frac{dy}{dx} - y}{x^2+y^2} &= \frac{x+y\frac{dy}{dx}}{x^2+y^2} \\[9pt] && \implies && x \frac{dy}{dx} - y &= x + y\frac{dy}{dx} \\ && \implies && (x-y) \frac{dy}{dx} &= x + y \\ && \implies && \frac{dy}{dx} &= \frac{x+y}{x-y}. \qquad \blacksquare \end{align*}$

Determine the derivative of a function g in terms of a function f

by RoRi

September 23, 2015

Determine the derivative of the function $g$ in terms of the function $f$ for the following definitions of $g$ :

$g(x) = f(x^2)$ ;
$g(x) = f(\sin^2 x) + f(\cos^2 x)$ ;
$g(x) = f(f(x))$ ;
$g(x) = f(f(f(x)))$ .

We use the chain rule to compute,
$g'(x) = (x^2)' f'(x^2) = 2x f'(x^2).$
We use the chain rule to compute,
$\begin{align*} g'(x) &= (\sin^2 x)' f'(\sin^2 x) + (\cos^2 x)' f'(\cos^2 x) \\ &= 2 \sin x \cos x f'(\sin^2 x) - 2 \sin x \cos x f'(\cos^2 x) \\ &= \sin (2x) f'(\sin^2 x) - \sin (2x) f'(\cos^2 x) \\ & = \sin (2x) (f'(\sin^2 x) - f'(\cos^2 x)). \end{align*}$
We use the chain rule to compute,
$g'(x) = (f'(f(x)))(f'(x)).$
Finally we use part (c) and the chain rule to compute,
$g'(x) = f'(f(f(x))) f'(x) f'(f(x)) = f'(x) f'(f(x)) f'(f(f(x))).$

Determine the value of g(x) = xf(x^2) and its derivatives given values of f(x)

by RoRi

September 23, 2015

Given a function $g(x) = xf(x^2)$ and the following table of values for $f$ and its derivatives:

$\begin{tabular}{|c | c | c | c |} \hline $x$ & $f(x)$ & $f'(x)$ & $f''(x)$ \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 1 & 1 \\ 2 & 3 & 2 & 1 \\ 4 & 6 & 3 & 0 \\ \hline \end{tabular}$

Compute the value of $g(x)$ , $g'(x)$ and $g''(x)$ for $x = 0,1,2$ .

First, by the chain rule we have

$\begin{align*} g'(x) &= f(x^2) + 2x^2 f'(x^2) \\ g''(x) &= 6xf'(x^2) + 4x^3 f''(x^2). \end{align*}$

Using these formulas we compute the table,

$\begin{tabular}{|c | c | c | c |} \hline $x$ & $g(x)$ & $g'(x)$ & $g''(x)$ \\ \hline 0 & 0 & 0 & 0 \\ 1 & 1 & 3 & 10 \\ 2 & 12 & 30 & 36 \\ \hline \end{tabular}$

Compute values of a composite function given values of the components

by RoRi

September 22, 2015

Given functions $f(x)$ and $g(x)$ with the following values

$\begin{tabular}{|c | c | c | c | c |} \hline $x$ & $f(x)$ & $f'(x)$ & $g(x)$ & $g'(x)$ \\ \hline 0 & 1 & 5 & 2 & -5 \\ 1 & 3 & -2 & 0 & 1 \\ 2 & 0 & 2 & 3 & 1 \\ 3 & 2 & 4 & 1 & -6 \\ \hline \end{tabular}$

Let $h(x) = f(g(x))$ and $k(x) = g(f(x))$ and construct a table as above for the functions $h$ and $k$ .

First, by the chain rule we have

$\begin{align*} h(x) &= f(g(x)) &\implies \quad h'(x) &= f'(g(x)) g'(x) \\ k(x) &= g(f(x)) &\implies \quad k'(x) &= g'(f(x)) f'(x). \end{align*}$

Using these formulas we compute the table,

$\begin{tabular}{|c | c | c | c | c |} \hline $x$ & $h(x)$ & $h'(x)$ & $k(x)$ & $k'(x)$ \\ \hline 0 & 0 & -10 & 0 & 5 \\ 1 & 1 & 5 & 1 & 12 \\ 2 & 2 & 4 & 2 & -10 \\ 3 & 3 & 12 & 3 & 4 \\ \hline \end{tabular}$

Compute the derivatives of two given functions

by RoRi

September 22, 2015

Consider the functions

$\begin{align*} f(x) &= \frac{1}{1 + \frac{1}{x}}, \qquad x \neq 0 \\ g(x) &= \frac{1}{1 + \frac{1}{f(x)}}. \end{align*}$

Find thd derivatives $f'(x)$ and $g'(x)$ .

First, we compute the derivative of $f$ ,

$\begin{align*} f'(x) &= \frac{\frac{1}{x^2}}{\left( 1+ \frac{1}{x} \right)^2} \\ &= \frac{1}{(x+1)^2}, \end{align*}$

where we multiplied the numerator and denominator by $x^2$ to get the second line.

Next, using the chain rule we know

$\left( 1+ \frac{1}{f(x)} \right)' = \frac{-f'(x)}{(f(x))^2}.$

So, then using the quotient rule and chain rule we can evaluate the derivative of $g$ ,

$\begin{align*} g'(x) &= \frac{\frac{f'(x)}{(f(x))^2}}{\left(1+ \frac{1}{f(x)}\right)^2} \\ &= \frac{f'(x)}{(f(x))^2 \left( 1 + \frac{1}{f(x)} \right)^2} \\ &= \frac{f'(x)}{(f(x) + 1)^2}. \end{align*}$

Then, plugging in the definition of $f(x)$ and the expression for $f'(x)$ which we already computed, we have,

$\begin{align*} g'(x) &= \frac{ \frac{1}{(x+1)^2} }{\left( \frac{1}{1+ \frac{1}{x}} + 1 \right)^2} \\[9pt] &= \frac{1}{(1+x)^2 \left( \frac{2x+1}{x+1} \right)^2} \\ &= \frac{1}{(2x+1)^2} \\ &= (2x+1)^{-2}. \end{align*}$

Compute the derivative of the given function

by RoRi

September 22, 2015

Compute the derivative of the function

$f(x) = (1+x)(2 + x^2)^{\frac{1}{2}} (3+x^3)^{\frac{1}{3}}$

where $x^3 \neq 3$ .

We compute this directly, (we can use this exercise to quickly get the formula for the derivative of a product of three differentiable functions, although it’s not really necessary since it is still a direct application of the product rule for derivatives)

$\begin{align*} f'(x) &= (1+x)' (2+x^2)^{\frac{1}{2}} (3+x^3)^{\frac{1}{3}} + (1+x)\left( (2+x^2)^{\frac{1}{2}} \right)' (3 + x^3)^{\frac{1}{3}} \\ & \qquad \qquad + (1+x)(2+x^2)^{\frac{1}{2}}\left( (3+x^3)^{\frac{1}{3}} \right)' \\ &= (2+x^2)^{\frac{1}{2}} (3+x^3)^{\frac{1}{3}} + (1+x)(3+x^3)^{\frac{1}{3}} \frac{x}{\sqrt{2+x^2}} + (1+x)(2+x^2)^{\frac{1}{2}} \frac{x^2}{(3+x^3)^{\frac{2}{3}}} \\[9pt] &= \frac{3x^5 + 2x^4 + 4x^3 + 8x^2 + 3x + 6}{(2+x^2)^{\frac{1}{2}} (3 + x^3)^{\frac{2}{3}}} \end{align*}$

Compute the derivative of the given function

by RoRi

September 22, 2015

Compute the derivative of the function

$\sqrt{x + \sqrt{x + \sqrt{x}}}.$

To help our computation, let’s define $g(x) = \sqrt{x + \sqrt{x}}$ . Then,

$\begin{align*} g'(x) &= \left( \frac{1}{2 \sqrt{x + \sqrt{x}}} \right) \left( 1 + \frac{1}{2 \sqrt{x}} \right) \\ &= \frac{1}{2 \sqrt{x + \sqrt{x}}} + \frac{1}{4 \sqrt{x} \sqrt{x + \sqrt{x}}} \\ & = \frac{1}{ 2 g(x)} + \frac{1}{(4 \sqrt{x}) g(x)}. \end{align*}$

Then, using this $g(x)$ and the chain rule, we have

$\begin{align*} f(x) &= \sqrt{x + g(x)} \\ \implies f'(x) &= \frac{1}{2 \sqrt{x + g(x)}} (1 + g'(x)) \\[9pt] &= \left( \frac{1}{2 \sqrt{x + g(x)}} \right) \left( 1 + \frac{1}{2 g(x)} + \frac{1}{ (4 \sqrt{x})g(x)} \right) \\[9pt] &= \frac{(4 \sqrt{x}) g(x) + 2 \sqrt{x} + 1}{(8 \sqrt{x}) g(x) \sqrt{x + g(x)}}. \end{align*}$

(This is the form the answer in the back of the book gives, so we’ll leave it in this form. If you want, you can plug back in $g(x) = \sqrt{x + \sqrt{x}}$ , but I don’t think it simplifies to anything nice.)

Compute the derivative of the given function

by RoRi

September 22, 2015

Compute the derivative of the function

$f(x) = \frac{1}{\sqrt{1+x^2} \left( x + \sqrt{1+x^2} \right)}.$

First, we simplify the expression for $f$ ,

$\begin{align*} f(x) &= \frac{1}{\sqrt{1+x^2} \left( \sqrt{1+x^2} + x \right)} \\[8pt] &= \frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2} \left( \sqrt{1+x^2} + x \right) \left( \sqrt{1+x^2} - x \right)} \\[8pt] &= \frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2} (1 + x^2 - x^2)} \\ &= 1 - \frac{x}{\sqrt{1+x^2}}. \end{align*}$

Now, we take the derivative. (Of course, you could take the derivative directly from the given form of $f$ , I just find this form somewhat easier, but it was not so obvious we could simplify $f$ this way.)

$\begin{align*} f'(x) &= - \left( \frac{ \sqrt{1+x^2} - \frac{1}{2} \frac{x}{\sqrt{1+x^2}} (2x) }{1+x^2}\right) \\[8pt] &= - \left( \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2}\right) \\[8pt] &= - \left( \frac{1+x^2 - x^2}{(1+x^2)^{\frac{3}{2}}}\right) \\[8pt] &= \frac{-1}{(1+x^2)^{\frac{3}{2}}}. \end{align*}$

Compute the derivative of the given function

by RoRi

September 21, 2015

Compute the derivative of the function

$f(x) = \left( \frac{1+x^3}{1-x^3} \right)^{\frac{1}{3}}.$

Using the chain rule we compute,

$\begin{align*} f'(x) &= \frac{1}{3} \left( \frac{1+x^3}{1-x^3} \right)^{-\frac{2}{3}} \left( \frac{3x^2(1-x^3) - (1+x^3)(-3x^2)}{(1-x^3)^2} \right) \\ &= \frac{1}{3} \left( \frac{1+x^3}{1-x^3} \right)^{-\frac{2}{3}} \left( \frac{3x^2 - 3x^5 + 3x^2 + 3x^5}{(1-x^3)^2} \right) \\ &= \frac{1}{3} \left( \frac{1+x^3}{1-x^3} \right)^{-\frac{2}{3}} \left( \frac{6x^2}{(1-x^3)^2} \right) \\ &= \frac{2x^2}{1-x^3} \frac{1}{1-x^3} \left( \frac{1+x^3}{1-x^3} \right)^{-\frac{2}{3}} \\ &= \frac{2x^2 (1+x^3)}{(1-x^3)(1+x^3)} \frac{1}{1-x^3} \left( \frac{1+x^3}{1-x^3} \right)^{-\frac{2}{3}} \\ &= \frac{2x^2}{1-x^6} \frac{1+x^3}{1-x^3} \left( \frac{1 + x^3}{1-x^3} \right)^{-\frac{2}{3}} \\ &= \frac{2x^2}{1-x^6} \left( \frac{1+x^3}{1-x^3} \right)^{\frac{1}{3}}. \end{align*}$

Compute the derivative of the given function

by RoRi

September 21, 2015

Compute the derivative of the function

$f(x) = \frac{x}{\sqrt{4-x^2}}.$

Using the chain rule we compute,

$\begin{align*} f'(x) &= \frac{\sqrt{4-x^2} - x \left(\frac{1}{2}\right) (-2x)(4-x^2)^{-\frac{1}{2}}}{4-x^2} \\[8pt] &= \frac{4-x^2+x^2}{(4-x^2)^{\frac{3}{2}}} \\[8pt] &= \frac{4}{(4-x^2)^{\frac{3}{2}}}. \end{align*}$