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Derivatives of inverse functions

1. Prove that ddx

arcsin(x) = 11x2 .

Proof: Set y = arcsin(x) so that we want to show dydx

= 11x2 . By

the definition of arcsine we know

x= sin(y) () and 2 y 2 ()

Differentiating both sides of () with respect to x gives

d

dx(x) =

d

dx(sin(y)) = 1 = d

dy(sin(y))

dy

dx

= 1 = cos(y)dy

dx =dy

dx =

1

cos(y) .

Now lets use the pythagorean identity:

sin2(y)+cos2(y) = 1 = cos2(y) = 1sin2(y) = cos(y) =

1 sin2(y)

By () we know cos(y) > 0 so that cos(y) =

1 sin2(y). So wehave

dy

dx=

1

cos(y)=

11 sin2(y)

()=

11 x2 .

2. Prove that ddx

arccos(x) = 11x2 .

Proof: Set y = arccos(x) so that we want to show dydx = 11x2 . Bythe definition of arccosine we know

x= cos(y) () and 0 y ()Differentiating both sides of () with respect to x gives

d

dx(x) =

d

dx(cos(y)) = 1 = d

dy(cos(y))

dy

dx

= 1 = sin(y)dydx

= dydx

= 1sin(y)

.

Now lets use the pythagorean identity:

sin2(y)+cos2(y) = 1 = sin2(y) = 1cos2(y) = sin(y) =

1 cos2(y)By () we know sin(y) > 0 so that sin(y) =

1 cos2(y). So we

havedy

dx= 1sin(y)

= 11 cos2(y)

()= 1

1 x2 .

7/23/2019 In Verses

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3. Prove that ddx

arctan(x) = 11+x2 .

Proof: Set y = arctan(x) so that we want to show dydx

= 11+x2

. By thedefinition of arctangent we know

x= tan(y) ()

[Note: it is also true that2 y 2 , but we will not need this factin the proof.] Differentiating both sides of () with respect to x gives

d

dx(x) =

d

dx(tan(y)) = 1 = d

dy(tan(y))

dy

dx

= 1 = sec2

(y)

dy

dx =dy

dx =

1

sec2(y) .

Now lets use the pythagorean identity:

sin2(y)+cos2(y) = 1 = sin2(y)

cos2(y)+

cos2(y)

cos2(y)=

1

cos2(y) = tan2(y)+1 = sec2(y)

So we have

dy

dx=

1

sec2(y) =

1

tan2(y) + 1

()=

1

1 +x2.

4. Prove that ddx

ln(x) = 1x

.

Proof: Set y = ln(x) so that we want to show dydx

= 1x

. By thedefinition of the natural logarithm we know

x= ey ()

Differentiating both sides of () with respect to x gives

d

dx(x) =

d

dx(ey) = 1 = d

dy(ey)

dy

dx = 1 =eydy

dx

= dydx

= 1

ey()

= dydx

= 1

x.

5. Prove that ddx

logb(x) = 1x ln(b) .

Proof: Set y = logb(x) so that we want to show dydx

= 1x ln(b) . By the

definition of a logarithm we know

x= by ()

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Differentiating both sides of () with respect to x gives

d

dx(x) =

d

dx(by) = 1 = d

dy(by)

dy

dx = 1 =by ln(b)dy

dx

= dydx

= 1

by ln(b)

()= dy

dx=

1

x ln(b).

6. Let fbe a differentiable function with inverse f1 which is also dif-ferentiable. Prove that

d

dxf1(x) =

1

f(f1(x))

as long as the denominator is not zero.

Proof: Set y = f1(x) so that we want to show dydx

= 1f(f1(x))

. By

the definition of inverse functions we know

x= f(y) ()

Differentiating both sides of () with respect to x gives

d

dx(x) =

d

dx(f(y)) = 1 = d

dy(f(y))

dy

dx = 1 =f(y)dy

dx

= dydx

= 1f(y)

= dydx

= 1f(f1(x))

.