Differentiation Techniques

1. Intuitive Limits

The limit of a function as x→0x \to 0, this function can be replaced so that calculating the limit is much simpler. The following replacements can be considered.

  • sin⁡x  ⟹  x\sin x \implies x,
    • sin⁡kx  ⟹  kx\sin kx \implies kx,   sin⁡nx  ⟹  xn\; \sin^n x \implies x^n
  • tan⁡x  ⟹  x\tan x \implies x,
    • tan⁡kx  ⟹  kx\tan kx \implies kx,   tan⁡nx  ⟹  xn\; \tan^n x \implies x^n
  • sin⁡−1x  ⟹  x\sin^{-1} x \implies x,
    • sin⁡−1kx  ⟹  kx\sin^{-1} kx \implies kx,   (sin⁡−1kx)n  ⟹  (kx)n\; (\sin^{-1} kx)^n \implies (kx)^n
  • tan⁡−1x  ⟹  x\tan^{-1} x \implies x,
    • tan⁡−1kx  ⟹  kx\tan^{-1} kx \implies kx,   (tan⁡−1kx)n  ⟹  (kx)n\; (\tan^{-1} kx)^n \implies (kx)^n
  • x3+x  ⟹  xx^3 + x \implies x.
    • x5−x3+x  ⟹  xx^5 - x^3 + x \implies x, so in this case addition/subtraction does not matter.

When xx is large enough, the following comparison is also useful. ∣sin⁡x∣<ln⁡x<xn<2x<x!<xx\begin{aligned} \vert \sin x \vert < \ln x < x^n < 2^x < x! < x^x \end{aligned}

2. Limit of Exponential Functions

There are a few ways to find the limits of the form of 000^0, ∞0\infty^0, or 1∞1^\infty.

  • Given y=f(x)g(x)y = f(x)^{g(x)}, taking a logarithm of both sides of the function may be helpful. In other words, the limit of yy can be found after calculating the limit of ln⁡y=g(x)ln⁡f(x)\ln y = g(x) \ln f(x).
  • The property ab=ebln⁡aa^b = e^{b \ln a} may be used.
{  lim⁡x→af(x)g(x)=lim⁡x→aeg(x)ln⁡f(x)  lim⁡x→af(x)1g(x)=lim⁡x→aeln⁡f(x)g(x)\begin{aligned} \begin{cases} \; \displaystyle \lim_{x \to a} f(x)^{g(x)} = \displaystyle \lim_{x \to a} e^{g(x) \ln {f(x)}} \\\\ \; \displaystyle \lim_{x \to a} f(x)^{\frac{1}{g(x)}} = \displaystyle \lim_{x \to a} e^{\frac{\ln {f(x)}}{g(x)}} \end{cases} \end{aligned}
  • Especially, in the case of 1∞1^\infty, the property ab=eb(a−1)a^b = e^{b (a-1)} may be used as well. Assuming that f(a)=1f(a) = 1 and g(a)=0g(a) = 0,
lim⁡x→aln⁡f(x)g(x)=lim⁡x→af′(x)f(x)g′(x)=lim⁡x→af′(x)g′(x)=lim⁡x→af(x)−f(a)x−ag(x)−g(a)x−a=lim⁡x→af(x)−1g(x)  ⟹  lim⁡x→af(x)1g(x)=lim⁡x→aeln⁡f(x)g(x)=lim⁡x→aef(x)−1g(x)\begin{aligned} \lim_{x \to a} \cfrac{\ln {f(x)}}{g(x)} = \lim_{x \to a} \cfrac{\cfrac{f'(x)}{f(x)}}{g'(x)} &= \lim_{x \to a} \cfrac{f'(x)}{g'(x)} = \lim_{x \to a} \cfrac{\cfrac{f(x) - f(a)}{x - a}}{\cfrac{g(x) - g(a)}{x - a}} = \lim_{x \to a} \cfrac{f(x) - 1}{g(x)} \\\\ \implies \lim_{x \to a} f(x)^{\frac{1}{g(x)}} &= \lim_{x \to a} e^{\frac{\ln {f(x)}}{g(x)}} = \lim_{x \to a} e^{\frac{f(x) - 1}{g(x)}} \end{aligned}
  • The following results are also useful.
{  lim⁡x→0(kx)hx=1  lim⁡x→∞(kx)hx=1,{  lim⁡x→0(kx)hx=1  lim⁡x→∞(kx)hx=1\begin{aligned} \begin{cases} \; \displaystyle \lim_{x \to 0} (kx)^{hx} = 1 \\\\ \; \displaystyle \lim_{x \to \infty} (kx)^{\frac{h}{x}} = 1 \end{cases}, \qquad \begin{cases} \; \displaystyle \lim_{x \to 0} \left(\cfrac{k}{x}\right)^{hx} = 1 \\\\ \; \displaystyle \lim_{x \to \infty} \left(\cfrac{k}{x}\right)^{\frac{h}{x}} = 1 \end{cases} \end{aligned}

3. Inverse Trigonometric Functions

Derivatives of inverse trigonometric functions are as follows.

  • (sin⁡−1x)′=11−x2(−π2≤sin⁡−1x≤π2)  (\sin^{-1} x)' = \cfrac{1}{\sqrt{1 - x^2}} \quad \left( -\cfrac{\pi}{2} \leq \sin^{-1} x \leq \cfrac{\pi}{2} \right) \;,
  • (cos⁡−1x)′=−11−x2(0≤cos⁡−1x≤π)  (\cos^{-1} x)' = \cfrac{-1}{\sqrt{1 - x^2}} \quad \left( 0 \leq \cos^{-1} x \leq \pi \right) \;,
  • (tan⁡−1x)′=11x2(−π2<sin⁡−1x<π2)  (\tan^{-1} x)' = \cfrac{1}{1 x^2} \quad \left( -\cfrac{\pi}{2} < \sin^{-1} x < \cfrac{\pi}{2} \right) \;,
  • (cot⁡−1x)′=−11+x2(0<cot⁡−1x<π)  (\cot^{-1} x)' = -\cfrac{1}{1 + x^2} \quad \left( 0 < \cot^{-1} x < \pi \right) \;,
  • (sec⁡−1x)′=1∣x∣x2−1(0≤sec⁡−1x≤π,  sec⁡−1x≠π2)  (\sec^{-1} x)' = \cfrac{1}{\vert x \vert \sqrt{x^2 - 1}} \quad \left( 0 \leq \sec^{-1} x \leq \pi, \; \sec^{-1} x \ne \cfrac{\pi}{2} \right) \;,
  • (csc⁡−1x)′=−1∣x∣x2−1(−π2≤csc⁡−1x≤π2,  csc⁡−1x≠0)  (\csc^{-1} x)' = -\cfrac{1}{\vert x \vert \sqrt{x^2 - 1}} \quad \left( -\cfrac{\pi}{2} \leq \csc^{-1} x \leq \cfrac{\pi}{2}, \; \csc^{-1} x \ne 0 \right) \;.

In addition, the following are important identities of inverse trigonometric functions.

  • sin⁡−1x+cos⁡−1x=π2  \sin^{-1}x + \cos^{-1}x = \cfrac{\pi}{2} \;,
  • sin⁡−1(−x)=−sin⁡−1x  \sin^{-1}(-x) = -\sin^{-1}x \;,
  • cos⁡−1(−x)=π−cos⁡−1x  \cos^{-1}(-x) = \pi - \cos^{-1}x \;,
  • csc⁡−1x=sin⁡−1(1x)  \csc^{-1} x = \sin^{-1}\left( \cfrac{1}{x} \right) \;.

4. Derivative of an Absolute Value Function

Usually, although this can be done by range, the following formula is noteworthy. (∣f(x)∣)′=f(x)f(x)′∣f(x)∣\begin{aligned} (\vert f(x) \vert)' = \cfrac{f(x) f(x)'}{\vert f(x) \vert} \end{aligned}

5. Derivative & Errors

By definition, the derivative of f(x)f(x) is as follows. lim⁡Δx→0ΔyΔx=f′(x)\begin{aligned} \lim_{\Delta x \to 0} \cfrac{\Delta y}{\Delta x} = f'(x) \end{aligned}

Since the average rate of change is not the same as the instantaneous rate of change in the view of the limit, the difference can be set as follows. ΔyΔx=f′(x)+ϵ\begin{aligned} \cfrac{\Delta y}{\Delta x} = f'(x) + \epsilon \end{aligned}

Note that ϵ\epsilon is smaller when Δx\Delta x gets smaller. It implies that the product of them is an infinitesimal number and it can be ignored. Therefore, the approximation of Δy\Delta y is f′(x)Δxf'(x) \Delta x. The limit of this approximation is called the derivative of yy and is written as dy=df(x)=f′(x)dx\begin{aligned} dy = df(x) = f'(x) dx \end{aligned}

Another point of view of dydy is that is is the error of yy. Moreover, dy/ydy/y is called the relative error. Meanwhile, this concept leads to the approximation formula about f(x+Δx)f(x + \Delta x). f(x+Δx)≈f(x)+f′(x)Δx\begin{aligned} f(x + \Delta x) \approx f(x) + f'(x) \Delta x \end{aligned}

6. Asymptotes

Asymptotes make it easier to draw a graph of a function. Vertical asymptotes can be found when a function goes to infinity as xx approaches a value. Also, horizontal asymptotes can be found when a function goes to a value as xx approaches ±∞\pm \infty. Other than these, there exist oblique asymptotes, which can be found in the following steps.

  • Divide both sides of the function by the highest order term and find a slope mm of the asymptote by evaluating the limits at infinity.
  • Set y=mx+ny = mx + n from the found mm and substitute this equation to the function. Again, divide both sides by the highest order term and find nn by evaluating the limits at infinity.

For example, the oblique asymptotes of y3=(x−1)(x−3)2y^3 = (x - 1)(x - 3)^2 can be found as follows. lim⁡x→∞(yx)3=lim⁡x→∞(x−1)(x−3)2x3=1  ⟹  y=x+n  ⟹  (x+n)3=(x−1)(x−3)2=x3−7x2+15x−9  ⟹  3nx2+3n2x+n3=−7x2+15x−9  ⟹  lim⁡x→∞3nx2+3n2x+n3x2=lim⁡x→∞−7x2+15x−9x2  ⟹  3n=−7  ⟹  y=x−73\begin{aligned} &\lim_{x \to \infty} \left( \cfrac{y}{x} \right)^3 = \lim_{x \to \infty} \cfrac{(x - 1)(x - 3)^2}{x^3} = 1 \\\\ &\implies y = x + n \\ &\implies (x + n)^3 = (x - 1)(x - 3)^2 = x^3 - 7x^2 + 15x - 9 \\ &\implies 3nx^2 + 3n^2x + n^3 = -7x^2 + 15x - 9 \\ &\implies \lim_{x \to \infty} \cfrac{3nx^2 + 3n^2x + n^3}{x^2} = \lim_{x \to \infty} \cfrac{-7x^2 + 15x - 9}{x^2} \\ &\implies 3n = -7 \\ &\implies y = x - \frac{7}{3} \end{aligned}

7. Curvature

Let y=f(x)y = f(x) be a twice-differentiable curve on a plane. Now, consider P(x,y)P(x, y) and Q(x+Δx,y+Δy)Q(x + \Delta x, y + \Delta y). When α\alpha is the angle between the tangent line on PP and xx-axis and α+Δα\alpha + \Delta \alpha is the angle between the tangent line on QQ and xx-axis, the tangent line is rotated by Δα\Delta \alpha moving the arc length Δs\Delta s on the curve. The mean curvature denotes the ratio of Δα\Delta \alpha to Δs\Delta s. Based on this concept, curvature KK of f(x)f(x) is computed as follows. K=∣y′′∣(1+(y′)2)32\begin{aligned} K = \cfrac{\vert y'' \vert}{(1 + (y')^2)^{\frac{3}{2}}} \end{aligned}

If this curve is a parametric function, KK is {  x=f(t)  y=g(t)  ⟹  K=∣x′⋅y′′−x′′⋅y′∣((x′)2+(y′)2)32\begin{aligned} \begin{cases} \; x = f(t) \\ \; y = g(t) \end{cases} \implies K = \cfrac{\vert x' \cdot y'' - x'' \cdot y' \vert}{((x')^2 + (y')^2)^{\frac{3}{2}}} \end{aligned}

Meanwhile, the radius of curvature RR can be calculated from KK. K=1R\begin{aligned} K = \cfrac{1}{R} \end{aligned}

Furthermore, the center of curvature (X,Y)(X, Y) on P(x,y)P(x, y) can be calculated as follows. X=x−y′(1+(y′)2)y′′,Y=y+1+(y′)2y′′\begin{aligned} X = x - \cfrac{y' (1 + (y')^2)}{y''}, \quad Y = y + \cfrac{1 + (y')^2}{y''} \end{aligned}

8. Gradient

The gradient has different definitions depending on whether the function is explicit or implicit. {  Explicit Function z=f(x,y)  ⟹  ∇f=∂f∂xi+∂f∂yj  Implicit Function f(x,y,z)=0  ⟹  ∇f=∂f∂xi+∂f∂yj+∂f∂zk\begin{aligned} \begin{cases} \; \text{Explicit Function } z = f(x, y) \implies \nabla f = \cfrac{\partial f}{\partial x} i + \cfrac{\partial f}{\partial y} j \\\\ \; \text{Implicit Function } f(x, y, z) = 0 \implies \nabla f = \cfrac{\partial f}{\partial x} i + \cfrac{\partial f}{\partial y} j + \cfrac{\partial f}{\partial z} k \end{cases} \end{aligned}

The gradient of an explicit function means the tangent vector at the point, while the gradient of an implicit function means the normal vector at the point. The details about this can be found in this note. Therefore, the normal vector from an explicit function can be obtained after reforming to an implicit function as follows. z−f(x,y)=0  ⟹  (−∂f∂x,  −∂f∂y,  1)\begin{aligned} z - f(x, y) = 0 \implies \left( -\cfrac{\partial f}{\partial x}, \; -\cfrac{\partial f}{\partial y}, \; 1 \right) \end{aligned}

9. Directional Derivative

The directional derivative of a multivariable differentiable function ff along a given vector uu at a given point xx represents the slope of the tangent at xx with the direction uu. If uu is parallel with the coordinate axes, the directional derivative is the partial derivative. Accordingly, the directional derivative can be viewed as an orthogonal projection of the partial derivative on uu. Consider P(a,b)P(a, b) on the function f(x,y)f(x, y). The directional derivative along the normalized direction uu at the point PP is Duf(P)=lim⁡h→0f(a+hux,b+huy)−f(a,b)h=∇f(P)⋅u=(∂f∂xi+∂f∂yj)⋅(uxi+uyj)=∂f∂xux+∂f∂yuy\begin{aligned} D_u f(P) &= \lim_{h \to 0} \cfrac{f(a + hu_x, b + hu_y) - f(a, b)}{h} = \nabla f(P) \cdot u \\\\ &= \left( \cfrac{\partial f}{\partial x} i + \cfrac{\partial f}{\partial y} j \right) \cdot (u_x i + u_y j) = \cfrac{\partial f}{\partial x}u_x + \cfrac{\partial f}{\partial y}u_y \end{aligned}

Therefore, the maximum of Duf(P)D_u f(P) is ∥∇f(P)∥\Vert \nabla f(P) \Vert and the minimum is −∥∇f(P)∥-\Vert \nabla f(P) \Vert.

10. Derivative of an Implicit Function

The general formula for the derivative of an implicit function is as follows. {  f(x,y)=0  ⟹  dydx=−fxfy  f(x,y,z)=0  ⟹  ∂z∂x=−fxfz,∂z∂y=−fyfz\begin{aligned} \begin{cases} \; f(x, y) = 0 \implies \cfrac{dy}{dx} = -\cfrac{f_x}{f_y} \\\\ \; f(x, y, z) = 0 \implies \cfrac{\partial z}{\partial x} = -\cfrac{f_x}{f_z}, \quad \cfrac{\partial z}{\partial y} = -\cfrac{f_y}{f_z} \end{cases} \end{aligned}

11. Bivariate Extreme Value

To find extreme values of a function f(x,y)f(x, y), the point (x,y)(x, y) such that fx=0f_x = 0 and fy=0f_y = 0 should be calculated. Let that point be (a,b)(a, b). If DD is defined as follows, (a,b)(a, b) is the extreme value when D>0D>0. D=fxx(a,b)fyy(a,b)−(fxy(a,b))2D>0  ⟹  {  fxx(a,b)>0local minimum  fxx(a,b)<0local maximum\begin{aligned} D &= f_{xx}(a, b) f_{yy}(a, b) - (f_{xy}(a, b))^2 \\\\ D > 0 &\implies \begin{cases} \; f_{xx}(a, b) > 0 & \text{local minimum} \\\\ \; f_{xx}(a, b) < 0 & \text{local maximum} \end{cases} \end{aligned}

If D<0D<0, there are no extreme values. Besides, if D=0D = 0, it is inconclusive.

12. Extrema of an Implicit Function

The extrema of yy of an implicit function f(x,y)=0f(x, y) = 0 can be calculated from fx=0f_x = 0 and f(x,y)=0f(x, y) = 0. Let such point be (a,b)(a, b). If fy≠0f_y \ne 0, {  −fxx(a,b)fy(a,b)>0local minimum b  −fxx(a,b)fy(a,b)<0local maximum b\begin{aligned} \begin{cases} \; -\cfrac{f_{xx}(a, b)}{f_y(a,b)} > 0 & \text{local minimum } b \\\\ \; -\cfrac{f_{xx}(a, b)}{f_y(a,b)} < 0 & \text{local maximum } b \end{cases} \end{aligned}

It is inconclusive if fxx(a,b)=0f_{xx}(a, b) = 0.

13. Lagrange Multiplier

When finding the maximum and minimum of f(x)f(x), this problem becomes often easy if an unknown variable can be represented using other unknown variables under constraints. Even so, however, the problem can sometimes get complex, and A is the proper method for this case. First, if a constraint g(x,y,z)=0g(x, y, z) = 0 is the only one, {  ∇f=λ∇g  g(x,y,z)=0\begin{aligned} \begin{cases} \; \nabla f = \lambda \nabla g \\\\ \; g(x, y, z) = 0 \end{cases} \end{aligned}

the maximum or minimum can be found by calculating λ\lambda and (x0,y0,z0)(x_0, y_0, z_0) satisfying the above conditions. Here λ\lambda is called the Lagrange multiplier. If another constraint h(x,y,z)=0h(x, y, z) = 0 is added, {  ∇f=λ∇g+μ∇h  g(x,y,z)=0  h(x,y,z)=0\begin{aligned} \begin{cases} \; \nabla f = \lambda \nabla g + \mu \nabla h \\\\ \; g(x, y, z) = 0 \\\\ \; h(x, y, z) = 0 \end{cases} \end{aligned}

the maximum or minimum can be found by calculating λ\lambda, μ\mu, and (x0,y0,z0)(x_0, y_0, z_0) satisfying the above conditions. For example, the intersection of a cylinder y2+z2=4y^2 + z^2 = 4 and a plane x+y+z=1x + y + z = 1 forms an ellipse. The Lagrange multiplier makes it simple to find the maximum and minimum of f=x+2yf = x + 2y for a point on this ellipse as follows. f=x+2y,{  g=x+y+z−1=0  h=y2+z2−4=0  ⟹  ∇f=(1,2,0),{  ∇g=(1,1,1)  ∇h=(0,2y,2z)  ⟹  ∇f=λ∇g+μ∇h  ⟺  (1,2,0)=λ(1,1,1)+μ(0,2y,2z)  ⟹  λ=1,λ+2yμ=2,λ+2zμ=0  ⟹  λ=1,y=12μ,z=−12μ  ⟹  h=y2+z2−4=14μ2+14μ2−4=0  ⟹  μ=±122,y=±22,z=∓22  ⟹  g=x+y+z−1=x−1=0  ⟹  x=1\begin{aligned} f &= x + 2y, \quad \begin{cases} \; g = x + y + z - 1 = 0 \\ \; h = y^2 + z^2 - 4 = 0 \end{cases} \\\\ &\implies \nabla f = (1, 2, 0), \quad \begin{cases} \; \nabla g = (1, 1, 1) \\ \; \nabla h = (0, 2y, 2z) \end{cases} \\\\ &\implies \nabla f = \lambda \nabla g + \mu \nabla h \iff (1, 2, 0) = \lambda(1, 1, 1) + \mu(0, 2y, 2z) \\ &\implies \lambda = 1, \quad \lambda + 2y\mu = 2, \quad \lambda + 2z\mu = 0 \\ &\implies \lambda = 1, \quad y = \cfrac{1}{2\mu}, \quad z = -\cfrac{1}{2\mu} \\ &\implies h = y^2 + z^2 - 4 = \cfrac{1}{4\mu^2} + \cfrac{1}{4\mu^2} - 4 = 0 \\ &\implies \mu = \pm \cfrac{1}{2\sqrt{2}}, \quad y = \pm 2\sqrt{2}, \quad z = \mp 2\sqrt{2} \\ &\implies g = x + y + z - 1 = x - 1 = 0 \\ &\implies x = 1 \end{aligned}

Therefore, the maximum is f(1,2,−2)=1+22f(1, \sqrt{2}, -\sqrt{2}) = 1 + 2\sqrt{2}, and the minimum is f(1,−2,2)=1−22f(1, -\sqrt{2}, \sqrt{2}) = 1 - 2\sqrt{2}.

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