Math Help?

[匿名]

2015-11-22 8:55 pm

I have a take home quiz for my Calculus class and i have just completed it; however i am unsure if my answers are correct. I tried to google answers to see if i did the problems correctly, but examples did not come up. Im just looking for the answers to the 2 questions to verify what i got. Thank you!

1) dy/dx of (e^6y)(cosx)=(5x^3)+(y^4)+(y)
2) equation of tangent line to f(x)=x/(csc(2x)+tan(x)) at x = pi/4.

回答 (1)

Lopez

2015-11-23 9:59 am

✔ 最佳答案

(1)
Let F = e^6y*cosx - 5x^3 - y^4 - y

dy/dx
= - Fx / Fy
= - ( - e^6y*sinx - 15x^2 ) / ( 6e^6y*cosx - 4y^3 - 1 )
= ( e^6y*sinx + 15x^2 ) / ( 6e^6y*cosx - 4y^3 - 1 ) ..... Ans

Method 2 :
d ( e^6y*cosx ) / dx = d ( 5x^3 + y^4 + y ) / dx
6e^6y*y ' *cosx - e^6y*sinx = 15x^2 + 4y^3y' + y'
y' * ( 6e^6y*cosx - 4y^3 - 1 ) = 15x^2 + e^6y*sinx
y' = ( 15x^2 + e^6y*sinx ) / ( 6e^6y*cosx - 4y^3 - 1 )

(2)
f '
= [ csc 2x + tan x - x( - 2csc 2x*cot 2x + sec^2 x ) ] / ( csc 2x + tan x )^2
= ( csc 2x + tan x + 2x*csc 2x*cot 2x - x*sec^2 x ) / ( csc 2x + tan x )^2

m
= f ' (π/4)
= [ 1 + 1 + (π/2)*1*0 - (π/4)(√2)^2 ] / (1+1)^2
= ( 2 - π/2 ) / 4
= ( 4 - π ) / 8

f (π/4)
= (π/4) / (1+1)
= π/8

the eq. of tangent line is :
y - f (π/4) = m ( x - π/4 )
y - π/8 = [ ( 4 - π ) / 8 ]( x - π/4 ) = [ ( 4 - π ) / 8 ]x - π/8 + π^2/32
y = [ ( 4 - π ) / 8 ]x + π^2/32 ..... Ans

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