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Prove or disprove given statements for functions such that f(x) = o(g(x))

by RoRi

Let f and g be functions, both differentiable in a neighborhood of 0, with g(x) > 0 and such that

    \[ f(x) = o(g(x)) \qquad \text{as} \qquad x \to 0. \]

Prove or disprove each the following statements.

  1. \displaystyle{ \int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right)} as x \to 0.
  2. f'(x) = o(g'(x)) as x \to 0.
  1. True.
    Proof. Since f(x) = o(g(x)) as x \to 0 we know by the definition of o(g(x)) that

        \[ \lim_{x \to 0} \frac{f(x)}{g(x)} = 0. \]

    Thus, for every \varepsilon > 0 there exists a \delta > 0 such that

        \[ |x| < \delta \quad \implies \quad \left| \frac{f(x)}{g(x)} \right| < \varepsilon. \]

    So, for |x| < \delta we have

        \begin{align*} \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| &\leq \frac{\int_0^x |f(t)| \, dt}{\left| \int_0^x g(t) \, dt \right|} \\[9pt] &< \frac{\varepsilon \int_0^x g(t) \, dt }{\left| \int_0^x g(t) \, dt \right|} \\[9pt] &= \varepsilon. \end{align*}

    The final line follows since g > 0 by hypothesis. Therefore,

        \[ |x| < \delta \quad \implies \quad \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| < \varepsilon. \]

    Hence,

        \[ \lim_{x \to 0} \frac{ \int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} = 0. \]

    By definition, we then have

        \[ \int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right). \qquad \blacksquare\]

  2. False.
    Consider f(x) = x^2 \sin \left( \frac{1}{x} \right) for x \neq 0 and f(x) = 0 for x = 0. Then, for x \neq 0,

        \[ f'(x) = 2x \sin \left( \frac{1}{x} \right) - \cos \left( \frac{1}{x} \right). \]

    For x = 0 we have f'(0) = 0.

    Next,

        \begin{align*} f(x) &= x^2 \sin \left( \frac{1}{x} \right) \\[9pt] &= x^2 \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right) \left( \frac{1}{x} \right) \\[9pt] &= x \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right). \end{align*}

    Since \lim_{x \to 0} \frac{\sin (1/x)}{1/x} = 1 we have f(x) = o(x) as x \to 0. However, f'(x) \neq o(1) since

        \[ \lim_{x \to 0} \left( 2 \sin \left( \frac{1}{x}\right) - \cos \left( \frac{1}{x} \right) \right) \]

    does not exist.

Prove an inequality of exponentials

by RoRi

For all x, y > 0 and for any constants a,b such that 0 < a < b prove that

    \[ \big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}. \]

Proof. We want to consider the function

    \[ f(t) = (x^t + y^t)^{\frac{1}{t}}. \]

If we can show this function is decreasing on the positive real axis then we establish the inequality since this would mean that if 0 < a < b then

    \[ f(b) < f(a) \quad \implies \quad (x^b+y^b)^{\frac{1}{b}} < (x^a+y^a)^{\frac{1}{a}}.\]

(So, the trick here is to think of this as a function of the exponent. The x and y are some positive fixed constants.) To take the derivative of f(t) we use logarithmic differentiation,

    \begin{align*} &&f(t) &= (x^t+y^t)^{\frac{1}{t}} \\[10pt] \implies &&\log f(t) &= \frac{1}{t} \log (x^t+y^t) \\[10pt] \implies &&\frac{d}{dt} (\log f(t)) &= \frac{d}{dt} \left( \frac{1}{t} \log (x^t+y^t) \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= -\frac{1}{t^2} \log (x^t+y^t) + \frac{1}{t} \left( \frac{1}{x^t+y^t} \right) \left( x^t \log x + y^t \log y \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t (x^t+y^t)}. \end{align*}

Multiplying both sides by f(t) we then obtain

    \begin{align*} f'(t) &= \left( \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t(x^t+y^t)} \right) (x^t+y^t)^{\frac{1}{t}} \\[10pt] &= (x^t + y^t)^{\frac{1}{t}} \left( \frac{t(x^t \log x + y^t \log y) - (x^t+y^t)\log(x^t+y^t)}{t^2 (x^t+y^t)} \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t t \log x + y^t t \log y - (x^t + y^t)\log(x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} \right) \left( x^t \log x^t + y^t \log y^t - x^t \log (x^t+y^t) - y^t \log (x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t \log \left(\frac{x^t}{x^t+y^t}\right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) \right). \end{align*}

Now we can conclude that f'(t) < 0 for all t > 0 since the first term in the product

    \[ \frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} > 0. \]

Since x, y > 0 (any real power of a positive number is still positive) and t^2 > 0. For the second term we have

    \[ x^t \log \left( \frac{x^t}{x^t+y^t} \right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) < 0 \]

since x^t and y^t are positive, but both logarithms are negative. We know these logarithms are negative since

    \[ \frac{x^t}{x^t+y^t} < 1 \quad \text{and} \quad \frac{y^t}{x^t+y^t} < 1 \]

implies

    \[ \log \left( \frac{x^t}{x^t+y^t} \right) < 0 \quad \text{and} \quad \log \left( \frac{y^t}{x^t+y^t} \right) < 0. \]

Hence, f'(t) < 0 for all t > 0. This means f(t) is a decreasing function. Therefore, if 0 < a < b then we have

    \[ f(b) < f(a) \quad \implies \quad \big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}. \qquad \blacksquare \]

Prove some inequalities of sin x

by RoRi

For all x > 0 prove that

    \[ x - \frac{x^3}{6} < \sin x < x. \]

From the first exercise of this section on inequalities, we know \sin x < x for all 0 < x < \frac{\pi}{2}. But since \sin x \leq 1 for all x and 1 < \frac{\pi}{2} we have the inequality on the right immediately,

    \[ \sin x < x \qquad \text{for all } x > 0. \]

For the inequality on the left let

    \[ f(x) = \sin x - x + \frac{x^3}{6}. \]

Then, we’ll consider the first two derivatives of f to show that it is positive for all x > 0.

    \begin{align*} f'(x) &= \cos x - 1 + \frac{x^2}{2} \\ f''(x) &= -\sin x + x = x - \sin x. \end{align*}

We know from the inequality on the right that \sin x < x for all x > 0. Hence, f''(x) > 0 for all x > 0. Therefore, f'(x) is increasing on the positive real axis. Since

    \[ f'(0) = \cos 0 - 1 + \frac{0^2}{2} = 0 \]

we then have that f'(x) > 0 for all x> 0. Hence, f(x) is increasing on the positive real axis, and since

    \[ f(0) = \sin 0 - 0 + \frac{0^3}{6} = 0 \]

we have that f(x) > 0 for all x >0. Thus, for all x>0,

    \[ \sin x - x + \frac{x^3}{6} > 0 \quad \implies \quad x - \frac{x^3}{6} < \sin x. \qquad \blacksquare \]

Prove the inequality 2x/π < sin x < x for 0 < x < π/2

by RoRi

Prove the inequality

    \[ \frac{2}{\pi}{x} < \sin x < x \qquad \text{for} \quad 0 < x < \frac{\pi}{2}. \]

Proof. Define a function f(x) = x - \sin x. Then, since \cos x < 1 for 0 < x < \frac{\pi}{2} we have

    \[ f'(x) = 1 - \cos x > 0 \qquad \text{for } 0 < x < \frac{\pi}{2}. \]

Since f(0) = 0 and f is increasing on 0 < x < \frac{\pi}{2} (since its derivative is positive) we have

    \[ f(x) > 0 \quad \implies \quad x - \sin x > 0 \quad \implies \quad x > \sin x. \]

For the inequality on the left (which is much more subtle), we want to show

    \[ \frac{2}{\pi}x < \sin x \qquad \text{which implies} \qquad frac{2}{\pi} < \frac{\sin x}{x}. \]

So, we consider the function

    \[ f(x) = \frac{\sin x}{x} \quad \implies \quad f'(x) = \frac{x \cos x - \sin x}{x^2}. \]

We know

    \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

and

    \[ f \left( \frac{\pi}{2} \right) = \frac{2}{\pi}. \]

Now, if we can show that f(x) is decreasing on the whole interval \left( 0, \frac{\pi}{2} \right) then we will be done (since this would mean f(x) > \frac{2}{\pi} on the whole interval since if it were less than \frac{2}{\pi} somewhere then it would have to increase to get back to \frac{2}{\pi} on the right end of the interval).

To show f(x) is decreasing on the whole interval we will show that its derivative is negative. To that end, define a function

    \[ g(x) = x \cos x - \sin x \]

(This is the numerator in the expression we got for f'(x). Since we know the denominator of that expression is always positive, we are going to show this g(x) is always negative to conclude f'(x) is always negative.) Now, g(0) = 0 and

    \[ g'(x) = \cos x - x \sin x - \cos x = -x \sin x < 0 \]

for 0 < x < \frac{\pi}{2}. Thus, g'(x) is negative, and so g(x) is decreasing. Since g(0) = 0 we then have g(x) is negative on the whole interval. Therefore, f'(x) \leq 0 on the interval. Hence, f(x) is decreasing. Hence, we indeed have

    \[ \frac{\sin x}{x} > \frac{2}{\pi} \quad \implies \quad \sin x > \frac{2}{\pi}x \]

for all x \in \left( 0, \frac{\pi}{2} \right). \qquad \blacksquare

Prove a formula for the 2nth derivative of x sin (ax)

by RoRi

Define f(x) = x \sin (ax) and prove the formula for the 2nth derivative of f,

    \[ f^{(2n)}(x) = (-1)^n (a^{2n} x \sin (ax) - 2n a^{2n-1} \cos (ax)). \]

Proof. The proof is by induction. For the case n = 1 we need to take two derivatives of f (since f^{(2n)} = f^{(2)} = f'').

    \begin{align*} && f(x) &= x \sin (ax) \\ \implies && f'(x) &= \sin (ax) + ax \cos (ax) \\ \implies && f''(x) &= a \cos (ax) + a \cos (ax) - a^2 x \sin (ax) \\ &&&= (-1)^n(a^{2n} x \sin (ax) - 2n a^{2n-1} \cos (ax)). \end{align*}

So, the formula holds for the case n = 1. Assume then that the formula is true for some positive integer k. Then the inductive hypothesis is that,

    \[ f^{(2k)}(x) = (-1)^k \left( a^{2k} x \sin (ax) - 2k a^{2k-1} \cos (ax) \right).\]

Now, we want to take two derivatives (to get a formula for f^{(2(k+1))}(x)).

    \begin{align*} f^{(2k+1)}(x) = \left(f^{(2k)}(x)\right)' &= \left( (-1)^k \left( a^{2k} x \sin (ax) - 2k a^{2k-1} \cos (ax)\right)\right)' \\[9pt] &= (-1)^k \big( a^{2k} \sin (ax) + a^{2k+1} x \cos (ax) + 2ka^{2k} \sin (ax) \big) \\[9pt] &= (-1)^k \big( (2k+1)a^{2k} \sin (ax) + a^{2k+1} x \cos (ax) \big). \end{align*}

Now, we take another derivative,

    \begin{align*} f^{(2(k+1))}(x) &= \left( f^{(2k+1)}(x)\right)' \\[9pt] &= \left( (-1)^k \big( (2k+1)a^{2k} \sin (ax) + a^{2k+1} x \cos (ax) \big) \right)' \\[9pt] &= (-1)^k \big( (2k+1)a^{2k+1} \cos (ax) + a^{2k+1} \cos (ax) - a^{2k+2} x \sin (ax) \big) \\[9pt] &= (-1)^k \big( -a^{2(k+1)} x \sin (ax) + (2k+2)a^{2k+1} \cos (ax) \big) \\[9pt] &= (-1)^{k+1} \big(a^{2(k+1)} x \sin (ax) - 2(k+1)a^{2(k+1) - 1} \cos (ax) \big). \end{align*}

Thus, the formula holds for k+1. Hence, it holds for all positive integers n. \qquad \blacksquare

Prove the formulas for derivatives of products and quotients

by RoRi

Derive the formulas for the derivative of a product and the derivative of a quotient from the corresponding formulas for the derivative of a sum and the derivative of a difference.

We know the derivative rules for sums and differences are:

    \[ \left( f(x) + g(x) \right)' = f'(x) + g'(x) \quad \text{and} \quad \left( f(x) - g(x) \right)' = f'(x) - g'(x). \]

To derive the derivative rule for products using logarithmic differentiation we let h(x) = f(x)g(x) and compute

    \begin{align*} h(x) = f(x)g(x) && \implies && \log |h(x)| &= \log |f(x)g(x)| \\[9pt] && \implies && (\log |h(x)|)' &= (\log |f(x)g(x)|)' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= (\log |f(x)| + \log |g(x)|)' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} \right) h(x) \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} \right) f(x)g(x) \\[9pt] && \implies && (f(x)g(x))' &= f'(x) g(x) + f(x)g'(x). \end{align*}

This is the usual rule for derivative of a product.

Similarly, for the derivative of a quotient, let h(x) = \frac{f(x)}{g(x)} and then compute,

    \begin{align*} h(x) = \frac{f(x)}{g(x)} && \implies && \log | h(x)| &= \log \left| \frac{f(x)}{g(x)} \right| \\[9pt] && \implies && (\log|h(x)|)' &= \left( \log \left| \frac{f(x)}{g(x)} \right| \right)' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= \left( \log |f(x)| - \log |g(x)| \right) ' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} \right) h(x) \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} \right) \frac{f(x)}{g(x)} \\[9pt] && \implies && \left( \frac{f(x)}{g(x)} \right)' &= \frac{f'(x)g(x) - g'(x)f(x)}{f(x)g(x)} \cdot \frac{f(x)}{g(x)} \\[9pt] && \implies && \left( \frac{f(x)}{g(x)} \right)' &= \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2}. \end{align*}

Which is the usual rule for derivative of a quotient.

Prove that a function satisfying given properties must be ex

by RoRi

Given a function f(x) satisfying the properties:

    \[ f(x+y) = f(x) f(y) \qquad \text{for all } x, y \in \mathbb{R}\]

and

    \[ f(x) = 1 + x g(x) \quad \text{where} \quad \lim_{x \to 0} g(x) = 1. \]

Prove the following:

  1. The derivative f'(x) exists for all x.
  2. We must have f(x) = e^x.

This problem is quite similar to two previous exercises here and here (Section 6.17, Exercises #39 and #40).

  1. Proof. To show that the derivative f'(x) exists for all x we must show that the limit

        \[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

    exists for all x. Using the given properties of f(x) we can evaluate this limit

        \begin{align*} \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} &= \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f(x)(hg(h))}{h} &(\text{using } f(h) = 1 + hg(h)) \\[9pt] &= \lim_{h \to 0} f(x)g(h) \\[9pt] &= f(x) &(\text{using } \lim_{h \to 0} g(h) = 1). \end{align*}

    Therefore, f'(x) = f(x) for all x, so the derivative is defined everywhere. \qquad \blacksquare

  2. Proof. From part (a) we know f(x) = f'(x). By Section 6.17, Exercise #39 (linked above) we know that the only functions f(x) which satisfy this equation are f(x) = 0 for all x or f(x) = Ke^x for some constant K (where c = 1 in the linked exercise). However, since the derivative of f exists everywhere, and differentiability implies continuity, we know f is continuous everywhere. Hence, \lim_{x \to 0} f(x) = f(0). Then,

        \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} (1 + xg(x)) = 1 + 0 = 1 \]

    since \lim_{x \to 0} g(x) = 1, so \lim_{x \to 0} xg(x) = 0. Therefore, we must have f(x) = Ke^x for some constant K. Furthermore, we must have K = 1 since f(0) = Ke^0 = K = 1. Thus, f(x) = e^x. \qquad \blacksquare

Find the slope and area under the graph for a given function

by RoRi

Let

    \[ f(x) = \sqrt{\frac{4x+2}{x(x+1)(x+2)}} \qquad \text{for} \quad x> 0. \]

  1. Determine the slope of the graph of f at the point with x-coordinate 1.
  2. Find the volume of the solid of revolution formed by rotating the region between the graph of f(x) and the interval [1,4] about the x-axis.
  1. To take this derivative, using logarithmic differentiation will be easier,

        \begin{align*} \log (f(x)) &= \log \left( \sqrt{ \frac{4x+2}{x(x+1)(x+2)}} \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log (x(x+1)(x+2)) \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log x - \log (x+1) - \log (x+2) \right). \end{align*}

    Then differentiating both sides we have,

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

    So, to find the slope at the point with x = 1 we evaluate,

        \[ f'(1) = \frac{1}{2} \left( \frac{2}{3} - 1 - \frac{1}{2} - \frac{1}{3} \right) = -\frac{7}{12}. \]

  2. First, the integral to compute the volume of the solid of revolution is,

        \begin{align*} V &= \pi \int_1^4 (f(x))^2 \, dx \\[9pt] &= \int_1^4 \frac{\pi(4x+2)}{x(x+1)(x+2)} \, dx. \end{align*}

    To evaluate this we use the partial fraction decomposition,

        \[ \frac{2x+1}{x(x+1)(x+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2}. \]

    This gives us the equation

        \[ A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) = 2x+1. \]

    Evaluating at x = 0, x = -1, and x = -2 we obtain

        \[ A = \frac{1}{2}, \quad B = 1, \quad C = -\frac{3}{2}. \]

    Therefore, we have

        \begin{align*} V &= 2 \pi \int_1^4 \left( \frac{1}{2x} + \frac{1}{x+1} - \frac{3}{2(x+2)} \right) \, dx \\[9pt] &= 2 \pi \left( \frac{1}{2} \int_1^4 \frac{1}{x} \,dx + \int_1^4 \frac{1}{x+1} \, dx - \frac{3}{2} \int_1^4 \frac{1}{x+2} \, dx \right) \\[9pt] &= \pi \log x \Bigr \rvert_1^4 + 2 \pi \log |x+1| \Bigr \rvert_1^4 - 3 \pi \log |x+2| \Bigr \rvert_1^4 \\[9pt] &= \pi \log 4 + 2 \pi (\log 5 - \log 2) - 3 \pi (\log 6 - \log 3) \\[9pt] &= 2 \pi \log 2 + 2 \pi \log 5 - 2 \pi \log 2 - 3 \pi \log (2 \cdot 3) + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 - 3 \pi \log 3 + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 \\[9pt] & = \pi (\log 25 - \log 8) \\[9pt] & = \pi \log \frac{25}{8}. \end{align*}

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

by RoRi

If

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

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*}