Find the derivative of the following functions: Write the solution of each number

Question

Find the derivative of the following functions: Write the solution of each number

5. y= ½tanh 2x + ⅔coth 3x

ans. y' = sech² 2x - 2csch23x

6. y = sinh2 (3x2 – 1)

ans. y' = 6xsinh(6x2 - 2)

7. f(z) = Arctan (sinh 6z)

ans. f'(z) = 6 sech 6z

Answer 1

Kevin,^'

There are a few basic tools [or Rules] that you could use in different combinations to solve each problem (A ,B below are functions of a variable x, A^' is dA(x)/dx etc.)

1) (A/B)^'= (A^' B-B^' A)/B²

2) A^'(cx)=cA^'(x) [c is a constant number ("coefficient")]

3) sinh^'(x)=cosh(x) ; cosh^'(x)=sinh(x)

4) A(B(x))'=A'(B(x))B^'(x)

5) Arctan(x)^'=1/1+x²

6) Use definitions of tanh(x), cotanh(x), sech(x) and cosech(x)

7) cosh²(x)-sinh²(x)=1

Try that out and that way you would be able to solve any problem like those

I'd show below as an example the step by step solution of pr #7 [they use z as a variable instead of x that I used for the Rules above]

f(z) = Arctan (sinh 6z) [For using the Rules above, use A would be the function Arctan, B - the function sinh , the coefficient c is 6

f(z)^'=[use Rules# 5,4,2]=1/(1+[sinh(6z)]²)*cosh(6z)*6=

[now use Rule #7]=[1/cosh²(6z)]*cosh(6z)*6=6/cosh(6z)=[Use Rule#6] =6sech(6z)

An Addendum: If you wonder where the rule #5 is coming from:

1. Atan(x)=tan^-1(x) [Atan is inverse of tan] (a)
2. Use the "Derivative of inverse f^-1(x)" rule

[f^-1(x)]^'=[1/f^'(y)]_f(y)=x (b)

as follows

Atan^'(x)=[1/tan^'(y)]_tan(y)=x (c)

Because tan^'(y)=1/cos²(y) (d)

[you could derive (d) for practice using some of the Rules 1-7 but modified to apply them for trigonometric (sin, cos etc) instead of hyperbolic ,(sh, ch etc.) functions]

Using a trigonometric identity

cos²(y)=1/[1+tan²(y)]

[again, for your practice, try to derive this one using definitions of cos and sin and Pythagorean theorem]

you'd get

1/cos²(y)=[1+tan²(y)] (e)

Therefore from (d) and (e)

tan^'(y)=1/cos²(y)=[1+tan²(y)] (f)

so from (b)

Arctan^'(x)=[1/tan^'(y)]_tan(y)=x=

{1/[1+tan²(y)]}_tan(y)=x=1/(1+x²) (g)

Hope this would be helpful and show you how to use problem solving to learn/review the material. This is one of the tools I use in my class teaching as well as in tutoring sessions.

Good luck Kevin,

Dr.Ariel B.