Question

This one is really hard I have been stuck for about 3 hours. PLEASE HELP

Answer 1

Given,

The expression is,

`√(y)+6xy=5`

Required

The double differentiation of the given function.

Differentiating the expression with respect to x then,

`\begin{gathered} (d)/(dx)√(y)+6xy=(d)/(dx)5 \\ (1)/(2√(y))(dy)/(dx)+6x(dy)/(dx)+6y=0 \\ (dy)/(dx)((1)/(2√(y))+6x)+6y=0 \end{gathered}`

Differentiating the function again with respect to x then,

`\begin{gathered} (d)/(dx)((dy)/(dx)((1)/(2√(y))+6x)+(d)/(dx)6y=0 \\ (d)/(dx)((1)/(2√(y))(dy)/(dx)+6x(dy)/(dx))+(d)/(dx)6y=0 \\ (1)/(2√(y))(d^2y)/(dx^2)-((dy)/(dx))^2(1)/(4y√(y))+6x(d^2y)/(dx^2)+6(dy)/(dx)+6(dy)/(dx)=0 \\ (d^2y)/(dx^2)((1)/(2√(y))+6x)=(1)/(4y√(y))((dy)/(dx))^2-12(dy)/(dx) \end{gathered}`

Substituting the value of dy/dx then,

`\begin{gathered} (d)/(dx)((dy)/(dx)((1)/(2√(y))+6x)+(d)/(dx)6y=0 \\ (d)/(dx)((1)/(2√(y))(dy)/(dx)+6x(dy)/(dx))+(d)/(dx)6y=0 \\ (1)/(2√(y))(d^2y)/(dx^2)-((dy)/(dx))^2(1)/(4y√(y))+6x(d^2y)/(dx^2)+6(dy)/(dx)+6(dy)/(dx)=0 \\ (d^2y)/(dx^2)((1)/(2√(y))+6x)=(1)/(4y√(y))((dy)/(dx))^2-12(dy)/(dx) \end{gathered}`