How do you properly apply the chain rule in implicit differentiation?

[Blurry__face14]

Homework Statement

Find the implicit differentiation

Relevant Equations

(sinx)^(cosy)+(cosx)^(siny)=a

The working I've tried is in the attachment.

 

[fresh_42]

I guess people do not want to download your picture, then rotate it, zoom in, only to find out that they cannot read it anyway.

Here is how to type formulas on PF (it's not difficult):
https://www.physicsforums.com/help/latexhelp/

 

[benorin]

I'm not going to follow your work, too taxing: I get this solution

$$\tfrac{dy}{dx}=\tfrac{( \cos x)^{\sin y}\sin y \tan x - ( \sin x)^{\cos y}\cos y \cot x}{( \cos x )^{ \sin y } \cos y \log \cos x - ( \sin x) ^{\cos y} \sin y \log \sin x}$$

 

[Blurry__face14]

I guess people do not want to download your picture, then rotate it, zoom in, only to find out that they cannot read it anyway.

Here is how to type formulas on PF (it's not difficult):
https://www.physicsforums.com/help/latexhelp/

Ahh, I apologise.
I've tried using Latex as you have asked, but I'm afraid it's taking way too long to type out my working.
However, I've taken a better photo, I'm not confident in my working but please do check. Thank you :)

 

[Blurry__face14]

I'm not going to follow your work, too taxing: I get this solution

$$\tfrac{dy}{dx}=\tfrac{( \cos x)^{\sin y}\sin y \tan x - ( \sin x)^{\cos y}\cos y \cot x}{( \cos x )^{ \sin y } \cos y \log \cos x - ( \sin x) ^{\cos y} \sin y \log \sin x}$$

Thank you for the answer. But may I ask what working you've done to solve this?

 

[benorin]

You forgot the chain rule when differentiating functions of y you need to multiply by y' from the chain rule, that'll give an equation involving y', solve it.