Chain and Inverse Function Rules

5.5 The Technique of Differentiation

5.5.3 Chain and Inverse Function Rules

DIFFERENTIATION TABLE

Line F(x) F⁰(x)≡(d/dx)F(x) Comments:

1 const 0

2 x^r rx^r−1 for the time being only r∈Z

3

4 sinx cosx -

5 cosx −sinx -

6 tanx 1/cos²x x6= (z+ 1/2)π, z ∈Z

7 cotx −1/sin²x x6=zπ, z ∈Z

8 arcsinx 9 arccosx 10 arctanx 11 arccotx

12 e^x e^x

13 r^x

14 ln|x|

15 log_b|x|

16 sinhx coshx

17 coshx sinhx

18 tanhx 1/cosh²x

19 cothx −1/sinh²x x6= 0

20 arsinhx 21 arcoshx 22 artanhx 23 arcothx

in Leibniz’s and in Lagrange’s notation:

(g(f(x)))⁰ =g⁰(y)f⁰(x).

Now, since we are familiar with the term differential this result may seem trivial to us, seeing how the fraction was simply reduced to higher terms bydy. Nevertheless, we want to quickly sketch theproof to demonstrate the advantages of the differentials, with whose help it is very simple:

First of all for the “inner” functiony=f(x) :dy=f⁰(x)dx+rf with lim

∆x→0rf(x,∆x)/dx= 0, then for the “outer” function z =g(y) :dz =g⁰(y)dy+rg with lim

∆y→0rg(y,∆y)/dy = 0.

After insertion it follows that: dz =g⁰(y)(f⁰(x)dx+r_f) +r_g =g⁰(y)f⁰(x)dx+g⁰(y)r_f+r_g, and after division by the differentialdx in the limit it becomes:

dz

dx ≡(_dx^d)g(f(x))≡g⁰(y)f⁰(x) = (_dy^dg)(_dx^df)≡(^dz_dy)(^dy_dx).

The followingexampleillustrates the advantages of the Leibniz notation: We are looking for the first derivative of ((x+ 1/x)⁴−1)³ forx6= 0 :

((x+ 1/x)⁴−1)³0

= (_dx^d)w(z(y(x)))

= (^dw_dz)(^dz_dy)(^dy_dx) according to the chain rule,

= 12 (x+ 1/x)⁴−12

(x+ 1/x)³(1−1/x²), since y=f(x) =x+ 1/x with (^dy_dx) = 1−1/x²,

z =g(y) =y⁴−1 with (_dy^dz) = 4y³ and w=h(z) =z³ with (^dw_dz) = 3z².

A further example is the

general exponential function: (b^x)⁰ =b^xlnb

Proof with y:=xlnb : (b^x)⁰ = (_dx^d)(e^xlnb) = (_dx^d)e^y = (_dy^d)e^y(_dx^dy) =e^ylnb=b^xlnb.

Exercise 5.3 Chain rule:

Calculate the following differential quotients using the chain rule:

a) (cosx)⁰ = (sin(^π₂ −x))⁰, b) (sinx²)⁰, c) (sin²x)⁰, d) (e^−x)⁰, e) (exp(−x²))⁰ and f ) _ax+b¹ 0

Finally we need the inverse function rule for the differential quotient of the inverse function x = f⁻¹(y) with y ∈ W_f of a differentiable bi-unique function y = f(x) with x ∈ D_f, whose differential quotient f⁰(x) =dy/dx 6= 0 is known and does not vanish in the wholeD_f:

inverse function rule: ^dx_dy = 1

(^dy_dx) for (^dy_dx)6= 0.

We want to derive this formula very simply in the Leibniz notation: In order to do this, we form the derivative ofx=f⁻¹(f(x)) with respect to xaccording to the chain rule:

1 = (_dx^d)(f⁻¹(f(x))) = (_dy^d)(f⁻¹(y))(^df(x)_dx ) = ^dx_dy^dy_dx and after division by ^dy_dx 6= 0 we arrive at the stated result.

Equipped with these rules we are able now to calculate all desired derivatives. Most of the proofs you will find in inserts:

First of all the

roots: y= ^m√

x=x^m1 for x >0 : ^m√

x⁰ = (x^m1)⁰ = (_m¹)x^m1−1 as inverse function of the exponential function x = y^m for y > 0, for ^m√

x⁰ = (x^1/m)⁰ = 1/(^dx_dy) = 1/my^m−1 = 1/m(x^1/m)^m−1 = (1/m)x^1/m−1,i.e. our power rule (*) holds also for reciprocal integers in the exponents.

Even more generally for

rational powers: z =x^mn forx >0 : (x^mn)⁰ = _mⁿx^mn−1

Meaning, ourpower rule (*) holds true even for any rational exponents.

Insert: Proof: with y=f(x) =x^1/m in the chain rule:

(x^n/m)⁰ = ( d

dx)((x^1/m)ⁿ) = dz dx = dz

dy ·dy dx

= (_dy^d)yⁿ·(_dx^d)x^1/m=nyⁿ⁻¹·(1/m)x^1/m−1

= (n/m)(x^1/m)ⁿ⁻¹x^1/m−1 = (n/m)xn/m−1/m+1/m−1

= (n/m)x^n/m−1

Then the

natural logarithm: y= lnx for x >0 : (lnx)⁰ = 1x for x6= 0 as inverse function to the exponential functionx=e^y fory ∈R.

Insert: Proof:

(lnx)⁰ = ^dy_dx = 1/(^dx_dy) = 1/(_dy^d)e^y = 1/e^y = 1/x for x6= 0. (Into the TABLE!)

Even the following holds true: (ln|x|)⁰ = 1x .

Because (ln−x)⁰ =dz/dx=dz/dy·dy/dx= 1/y(−1) =−1/(−x) = 1/xfor x6= 0.

We then turn our attention to the

general power: z =x^r =e^rlnx with r∈R: (x^r)⁰ =rx^r−1

i.e. our power rule (*) holds universally true even for any real exponent.

Insert: Proof: With z = e^y and y = rlnx in the chain rule we get: (x^r)⁰ =

dz

dx = ^dz_dy·_dx^dy =e^yr/x= (r/x)e^rlnx= (r/x)x^r=rx^r−1.(Into the TABLE: L.2!) Even for the

general logarithm: y= log_bx for x >0 : (log_bx)⁰ = 1 xlnb

to any real base b ∈ R, we now obtain the derivative, namely as inverse function of the general exponential function x=b^y :

Insert: Proof:

(log_bx)⁰ = ^dy_dx = 1

(^dx_dy) = 1

b^ylnb = 1

xlnb. (Into the TABLE: L.15!)

We conclude this list of differential quotients, which is important also for he following chapters, with the cyclometric and the area functions:

For the cyclometric functions, the inverse functions to the trigonometric ones, we get

arc tangent: for −π/2<arctanx < π/2 : (arctanx)⁰ = 1 1 +x²

Insert: Proof: With the inverse function x = tany, where from ^dx_dy = 1/cos²y = (cos²y + sin²y)/cos²y = 1 + tan²y = 1 +x² follows: (arctanx)⁰ =

dy

dx = 1/(^dx_dy) = 1/(1 +x²). (Into the TABLE: L.10!) Analogously for the

arc cotangent for 0<arccotx < π: (arccotx)⁰ =− 1 1 +x²

Exercise 5.4 Prove this with the inverse function: x= coty.

For the

arc cosine for−π/2<arcsinx < π/2 : (arcsinx)⁰ = √ 1

1−x² for |x|<1

Insert: Proof: With the inverse function x = siny, where from ^dx_dy = cosy = p(1−sin²y) = p

(1−x²) for |x| < 1 follows: (arcsinx)⁰ = ^dy_dx = 1/(^dx_dy) = 1/p

(1−x²). (Into the TABLE: L.8!) Analogously for

arc cosine for 0<arccosx < π: (arccosx)⁰ =− 1

p(1−x²) for |x|<1

Exercise 5.5 Prove this with the inverse function: x= cosy.

The area functions the inverse functions of the hyperbolic functions, complete our dif- ferentiation table:

For the

area hyperbolic tangent: (artanhx)⁰ = 1

1−x²for |x|<1

and the

area hyperbolic cotangent: (arcothx)⁰ =− 1

x²−1 for |x|>1.

Exercise 5.6 Prove this with the inverse function: x = tanhy, respectively with x = cothy.

For the

area hyperbolic sine: (arsinhx)⁰ = 1

p(1 +x²) for x∈R.

and

area hyperbolic cosine: 0<arcoshx: (arcoshx)⁰ = 1

p(x²−1) for x≥1.

Exercise 5.7 Prove this with the inverse function: x = sinhy, respectively with x = coshy ≥1, bi-unique only for y >0.

You will find all the preceding results combined below results in the bigdifferentiation table, to which we will return very often later on:

DIFFERENTIATION TABLE

Line F(x) F⁰(x)≡(d/dx)F(x) Comments:

1 const 0

2 x^r rx^r−1 r∈R

3

4 sinx cosx

5 cosx −sinx

6 tanx 1/cos²x x6= (z+ 1/2)π, z ∈Z

7 cotx −1/sin²x x6=zπ, z ∈Z

8 −π/2<arcsinx < π/2 1/p

(1−x²) |x|<1 9 0<arccosx < π −1/p

(1−x²) |x|<1 10 −π/2<arctanx < π/2 1/(1 +x²)

11 0<arccotx < π −1/(1 +x²)

12 e^x e^x

13 r^x r^xlnr 0< r∈R

14 ln|x| 1/x x6= 0

15 log_b|x| 1/xlnb x6= 0, 0< b∈R, b6= 1

16 sinhx coshx

17 coshx sinhx

18 tanhx 1/cosh²x

19 cothx −1/sinh²x x6= 0

20 arsinhx 1/p

(x²+ 1)

21 0<arcoshx 1/p

(x²−1) x >1

22 artanhx 1/(1−x²) |x|<1

23 arcothx −1/(x²−1) |x|>1

Exercise 5.8 Differentiation examples

Determine the differential quotients for the following functions y = f(x) with constants a, b, c and d:

a) y= sin³(4x), b) y= exp(−(x/a)²), c) y = ^{√ 1}

ax²+b, d) y= ln(3e^2x), e) y=acosh^x−b_a , f ) y =ax²exp(−bx), g) y= cos(ax+b) sin(cx+d), h) y= _1+(x/a)¹ 2, i) y=

sin(x/a) (x/a)

2

, j) y= arctan(1/x) + (x/2) (lnx²−ln(x²+ 1)) Calculate the first five derivatives of the following functions f(x) which we will need in the next chapter:

k) f(x) = sinx, l) f(x) = tanx, m) f(x) = e^x and n) f(x) = _1−x¹2