Question

Don’t get part b at all. Please may you explain it. I’m so stressed out I need help with part b only

Answer 1

Given the initial expression,

`x=3\tan 2y`

Then,

`(d)/(dx)(x)=(d)/(dx)(3\tan (2y))`

Thus,

`\begin{gathered} \Rightarrow(dx)/(dx)=1 \\ \Rightarrow1=(d)/(dx)(3\tan (2y)) \\ \Rightarrow1=3(d)/(dx)(\tan 2y) \\ \Rightarrow1=3(\sec ^22y\cdot2d(y)/(dx))\to\text{chain rule} \end{gathered}`

Thus,

`\begin{gathered} \Rightarrow1=6\sec ^22y\cdot(dy)/(dx) \\ \Rightarrow(1)/(6\sec^22y)=(dy)/(dx) \end{gathered}`

Using the trigonometric identity,

`1+\tan ^2z=\sec ^2z`

Then,

`\Rightarrow(1)/(6(1+\tan^22y))=d(y)/(dx)`

From the initial equation,

`\begin{gathered} x=3\tan 2y \\ \Rightarrow(x)/(3)=\tan 2y \end{gathered}`

Finally,

`\begin{gathered} \Rightarrow(1)/(6(1+\tan^22y))=(1)/(6(1+((x)/(3))^2))=d(y)/(dx) \\ \Rightarrow d(y)/(dx)=(1)/(6(1+((x)/(3))^2)) \end{gathered}`

Simplifying,

`\begin{gathered} \Rightarrow(dy)/(dx)=(1)/(6(1+(x^2)/(9)))=(1)/((6)/(9)(9+x^2))=(1)/((2)/(3)(9+x^2))=(3)/(2(9+x^2)) \\ \Rightarrow(dy)/(dx)=(3)/(18+2x^2) \end{gathered}`

The answer is dy/dx=3/(18+2x^2)