IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

nishant

let y=sinx,then d^2(cos^7x)/dy^2=

vinu

Hi nishant,

d²(cos⁷x)/dy² =(d/dy){d(1-y²)^7/2 /dy} ; [writing cosx in terms of y]

= -7{d(y[1-y²]^5/2 ) / dy} ;

= -7(1-y²)^3/2(1-6y²) .

nishant

nope.u r not even near to the answer....

vinu

Hey nishant,
check out my edited solution.
Hope it's correct.

magiclko

hey here is one solution

given that y=sin x

dy/dx = cos x

dx/dy = sec x

now let t= cos ⁷ x

then d (cos ⁷ x)/ dy = (dt/dx) (dy/dx)

= (-7 cos ⁶ x sin x) (sec x)

= -7 cos ⁵ x sin x

now let u = -7 cos ⁵ x sin x

thrfore d² (cos ⁷ x) / dy² = du/ dy

= (du/dx) (dx/dy)

= (-7cos ⁶ x + 35 cos⁴ x sin² x) (sec x)

= -7cos ⁵ x + 35 cos³ x sin² x

pls tell if m wrong anywhere!!!

nishant

there is no sine term in the answer,magicklo.vinu,ur answer too is wrong...

nachiket

y=sinx

(cosx)^7=(1-y^2)^7/2=let=a

da/dy=7/2(1-y^2)^5/2.-2y=-7y(1-y^2)^5/2

d^2a/dy=-7{(1-y^2)^5/2+y(5/2(1-y^2)^3/2.-2y)}=-7{(cosx)^5+5/2.(cosx)^3.-2(sinx)^2}=-7(cosx)^3{(cosx)^2-10(sinx)^2}

some more simplifications and we have the answer

peter_pete

Is the answer 7(1-y^2)^3/2(11y^2-1)

nishant

both of u r wrong...

KAB

Is this correct -42cos^5x+35cos^3x

nrki99

is the correct answer is 42 cos^3x -49 cos^5x

magiclko

arey ........ get sin² x = 1-cos² x ....

nishant

kab ur answer is 90 percent correct.could u just post ur soln. so that i can rectify ur mistake..

vinu

Hey nishant,
mine,magiclko, and kab' s answers are one & same.
substitute.....y=sinx in my ans ;
substitute.....sin^2 x =1-cos^2 x in it ,magiclko's ans.......
u get kab's ans .
Oh ! and i'm damn sure it's correct!!!
plz....tell me what ans u've got???

Ask & Discuss Questions with Community & Experts