Question

y =sin²(x⁴),find dy/dx.​

Answer 1

Answer:

Hope this helps you

Step-by-step explanation:

Let p=x2x4+a4−−−−−−√−−(1)p=x2x4+a4−−(1)

Substitute x2=a2tan(θ)x2=a2tan⁡(θ)

=>p=x2x4+a4−−−−−−√=>p=x2x4+a4

=>p=a2tan(θ)a4tan2(θ)+a4−−−−−−−−−−−−√=>p=a2tan⁡(θ)a4tan2⁡(θ)+a4

=>p=a2tan(θ)a2sec(θ)=sin(θ)=>p=a2tan⁡(θ)a2sec⁡(θ)=sin⁡(θ)

y=sin−1(x2x4+a4−−−−−−√)=sin−1

=>y=θ=>y=θ

=>dydθ=1−−(2)=>dydθ=1−−(2)

x2=a2tan(θ)x2=a2tan⁡(θ)

2xdxdθ=a2sec2(θ)=a2(1+tan2(θ))2xdxdθ=a2sec2⁡(θ)=a2(1+tan2⁡(θ))

=a2+a2tan2(θ)=a2+x4a2=a4+x4a2−−(3)=a2+a2tan2⁡(θ)=a2+x4a2=a4+x4a2−−(3)

dxdθ=a4+x42a2xdxdθ=a4+x42a2x

=>dydx=dydθ÷dxdθ