Question

then
show
that nou tydu - sinzu
27​

Answer 1

Step-by-step explanation:

u = tan-¹(x³ + y³/x - y)

tanu = x³ + y³ / x - y

= x³(1+y³/x³)/x(1-y/x)

= x²(1+y³/x³)/(1-y/x)

अतः u , 2 घात का समघात फलन है ।

x du/dx + y du/dy = nu

x d/dx(tanu) + y d/dy(tanu) = 2tanu

xsec²u du/dx + ysec²u du/dy = 2tanu

sec²u(x du/dx + y du/dy) = 2tanu

x du/dx + y du/dy = 2tanu/sec²u

= 2*(sinu/cosu)*cos²u

= 2sinu*cosu

= sin2u