# if y=sinx , then show d^2(cos^7x)/dy^2 is equal to 35cos^3x – 42cos^5x

Rinkoo Gupta
9 years ago
y=sinx
so dy/dx =cosx
let z=cos7x
dz/dy=7cos6x.(-sinx).dx/dy
=-7cos6xsinx.(1/cosx)
=-7cos5xsinx
d2z/dy2=d/dy(dz/dy)
=d/dx(dz/dy).dx/dy
=d/dx(-7cos5xsinx).(1/cosx)
=-7{cos5x.cosx+sinx.(-5cos4xsinx)}.(1/cosx)
=-7cos5x+35cos3xsin2x
=35cos3x(1-cos2x)-7cos5x
=35cos3x-35cos5x-7cos5x
=35cos3x-42cos5x

Thanks & Regards
Rinkoo Gupta
Arun Kumar IIT Delhi
9 years ago
Hi
$\\z=cos^7(x) \\y=sin(x) \\\frac{d^2z}{dy^2}=? \\\frac{dz}{dy}=\frac{\frac{dz}{dx}}{\frac{dy}{dx}}=\frac{7cos^6x*(-sinx)}{cosx} \\\frac{dz}{dy}=7cos^5x*(-sinx) \\\frac{d^2z}{dy^2}=\frac{d\frac{dz}{dy}}{dy}=\frac{d\frac{dz}{dy}}{dx}*\frac{1}{\frac{dy}{dx}} \\=\frac{-7\cos \left(x\right)\left(\cos ^5\left(x\right)-5\cos ^3\left(x\right)\sin ^2\left(x\right)\right)}{cos(x)} \\=>\frac{-7\cos \left(x\right)\left(\cos ^5\left(x\right)-5\cos ^3\left(x\right)\(1-cos^2(x))\left(x\right)\right)}{cos(x)}$
Solving last step should get you the answer.
Arun Kumar , IIT Delhi, Askiitians Faculty
9 years ago
f=cos7x and y=sinx df/dx=-7cos6xsinx dy/dx=cosx
so df/dy=-7cos5xsinx
now d(df/dy)/dx=35cos4xsin2x-7cos6x
so d2f/dy2=35cos3xsin2x-7cos5x
=35cos3x-42cos5x {df/dy=(df/dx)/(dy/dx)}