Further Differentiation Calculus?

Question

If y=e^xtanx, show that d²y/dx² = e^x(2tan³x + 2tan²x + 3tanx +2)

10pts given to first correct answer with clear working thanks :)

Answer 1

First, the needed differentiation rules:

Product rule for differentiation:

d/dx [ g(x) · h(x) ] = g(x) · d/dx [ h(x) ] + h(x) · d/dx [ g(x) ]

or, written more simply:

( a·b )' = a·b' + b·a'

Derivative of tangent:

d/dx [ tan(x) ] = sec²(x)

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y = e^(x) · tan(x)

Differentiate twice with respect to x:

d/dx [ y ] = d/dx [ e^(x) · tan(x) ]

You have to use the product rule on the right, since there are two functions of x multiplied together:

dy/dx = e^(x) · d/dx [ tan(x) ] + tan(x) · d/dx [ e^(x) ]

dy/dx = e^(x) · sec²(x) + tan(x) · e^(x)

——————————————————————————————————————

Use the Pythagorean trigonometric identity:

sin²θ + cos²θ = 1

sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ

tan²θ + 1 = sec²θ

——————————————————————————————————————

Replace secant squared and factor out e^x:

dy/dx = e^(x) · ( tan²(x) + 1 ) + tan(x) · e^(x)

dy/dx = e^(x) · ( tan²(x) + tan(x) + 1 )

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Differentiate again:

d/dx [ dy/dx ] = d/dx [ e^(x) · ( tan²(x) + tan(x) + 1 ) ]

Use the product rule:

d²y/dx² = e^(x) · d/dx [ ( tan²(x) + tan(x) + 1 ) ] + ( tan²(x) + tan(x) + 1 ) · d/dx [ e^(x) ]

Differentiate and use the chain rule on tangent squared:

= e^(x)·( 2·tan(x)·d/dx[ tan(x) ] + sec²(x) + 0 ) + ( tan²(x) + tan(x) + 1 )·e^(x)

= e^(x)·( 2·tan(x)·sec²(x) + sec²(x) ) + ( tan²(x) + tan(x) + 1 )·e^(x)

Factor out secant:

= e^(x)·{ [ 2·tan(x) + 1 ]·sec²(x) } + ( tan²(x) + tan(x) + 1 )·e^(x)

Use the identity to change secant to tangent again:

= e^(x)·{ [ 2·tan(x) + 1 ]·[ tan²(x) + 1 ] } + ( tan²(x) + tan(x) + 1 )·e^(x)

Use FOIL:

= e^(x)·[ 2·tan³(x) + 2·tan(x) + tan²(x) + 1 ] + ( tan²(x) + tan(x) + 1 )·e^(x)

Factor out e^(x) and combine like terms:

= e^(x)·[ 2·tan³(x)  + 2·tan²(x) + 3·tan(x) + 2 ]

Answer 2

step 1. (uv)' = uv' + u'v

where the little ' means derivative.

we also know that:

d tan x/dx = sec ^2 x

and

d sec x / dx = sec x tan x

and

d sec^2 x/dx = 2 sec^2 x tan x

dy/dx = e^x * sec^2x + e^x * tan x

step 2. again

e^x *  sec^2x  + e^x * 2 sec^2 x tan x + e^x * sec^2 x + e^x * tan x

now simplify:

e^x * (sec^2 x + 2 sec^2 x tan x +  sec^2 x + tan x)

We also know  from trigonometry that sec^2 x = 1 + tan^2 x

e^x * (sec^2 x + 2 sec^2 x tan x +  sec^2 x + tan x) becomes

e^x * ( 1 + tan^2 x + 2 sec^2 x tan x +  1 + tan^2 x + tan x)

and

e^x * ( 1 + tan^2 x + 2 tan ^3 x + 2 tan x  +  1 + tan^2 x + tan x)

simplified:

e^x * (2 tan ^3 x  + 2  tan^2 x + 3tan x + 2)

Please check my calculations.

Answer 3

y=e^xtanx

dy/dx=e^xtanx+e^xsecc

=e^x(tanx+sec²x)

By using chain rule to diferentiate sec²x

=secx(secxtanx)+secx(secxtanx)

=2secx(secxtanx)

=2tanxsec²x

=2tanx(1+tan²x)

=2tanx+2tan³x

use chain rule to direfentiate dy/dx

d²y/dx²=e^x(tanx+sec²x)+e^x(sec²x+2tan...

=e^x(tanx+1+tan²x)+e^x(1+tan²x+2tanx+2...

=e^x(tanx+1+tan²x+1+tan²x+2tanx+2tan³x...

=e^x(2tan³x+2tan²x+3tanx+2)

Answer 4

y=e^xtanx,

Use the product rule

dy/dx = e^x sec^2(x) + e^x tan(x)

d^2y/dx^2 = use the product rule for each term.

d^2y/dx^2 = e^x sec^2(x)+ e^x 2 sec(x) sec(x) tan(x) + e^x sec^2(x) + e^x tan(x)

d^2y/dx^2 = 2 e^xsec^2(x) + e^xtan(x) [ 2sec^2(x) +1]