How to do this question

[mimie]

I attempt to find the gradient first,
[MATH] \frac{d}{dx}\left[ \sin \left( x-y \right) \right]=\frac{d}{dx}\left[ x\cos \text{ (}y+\frac{\pi }{4}\text{)} \right] \\ \cos \left( x-y \right)\cdot \left( 1-\frac{dy}{dx} \right)=\cos \text{ (}y+\frac{\pi }{4}\text{)}-x\sin \text{ (}y+\frac{\pi }{4}\text{)}\frac{dy}{dx} \\ x\sin \text{ (}y+\dfrac{\pi }{4}\text{)}\dfrac{dy}{dx}-\cos \left( x-y \right)\dfrac{dy}{dx}=\cos \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right) \\ \left[ x\sin \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right) \right]\dfrac{dy}{dx}=\cos \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right) \\ \dfrac{dy}{dx}=\dfrac{\cos \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right)}{x\sin \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right)} \\ \text{At (}\dfrac{\pi }{4},\dfrac{\pi }{4}),\,\,\,\,\,\,\,\,\dfrac{dy}{dx}=\dfrac{\cos \text{ (}\dfrac{\pi }{4}+\dfrac{\pi }{4}\text{)}-\cos \text{(}\dfrac{\pi }{4}-\dfrac{\pi }{4}\text{)}}{\dfrac{\pi }{4}\sin \text{ (}\dfrac{\pi }{4}+\dfrac{\pi }{4}\text{)}-\cos \text{(}\dfrac{\pi }{4}-\dfrac{\pi }{4}\text{)}} \\ =\dfrac{\cos \text{ }\dfrac{\pi }{2}-\cos \text{ }0}{\dfrac{\pi }{4}\sin \text{ }\dfrac{\pi }{2}-\cos 0} \\ =\dfrac{-1}{\dfrac{\pi }{4}-1} \\ =\dfrac{-1}{\text{(}\dfrac{\pi -4}{4}\text{)}} \\ =\dfrac{-4}{\pi -4} \\ =\dfrac{4}{4-\pi } [/MATH]
Not sure if my work correct or not. Can someone check my calculation? Thanks in advance.

 

[Deleted member 4993]

View attachment 21023

I attempt to find the gradient first,
[MATH] \frac{d}{dx}\left[ \sin \left( x-y \right) \right]=\frac{d}{dx}\left[ x\cos \text{ (}y+\frac{\pi }{4}\text{)} \right] \\ \cos \left( x-y \right)\cdot \left( 1-\frac{dy}{dx} \right)=\cos \text{ (}y+\frac{\pi }{4}\text{)}-x\sin \text{ (}y+\frac{\pi }{4}\text{)}\frac{dy}{dx} \\ x\sin \text{ (}y+\dfrac{\pi }{4}\text{)}\dfrac{dy}{dx}-\cos \left( x-y \right)\dfrac{dy}{dx}=\cos \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right) \\ \left[ x\sin \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right) \right]\dfrac{dy}{dx}=\cos \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right) \\ \dfrac{dy}{dx}=\dfrac{\cos \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right)}{x\sin \text{ (}y+\dfrac{\pi }{4}\text{)}-\cos \left( x-y \right)} \\ \text{At (}\dfrac{\pi }{4},\dfrac{\pi }{4}),\,\,\,\,\,\,\,\,\dfrac{dy}{dx}=\dfrac{\cos \text{ (}\dfrac{\pi }{4}+\dfrac{\pi }{4}\text{)}-\cos \text{(}\dfrac{\pi }{4}-\dfrac{\pi }{4}\text{)}}{\dfrac{\pi }{4}\sin \text{ (}\dfrac{\pi }{4}+\dfrac{\pi }{4}\text{)}-\cos \text{(}\dfrac{\pi }{4}-\dfrac{\pi }{4}\text{)}} \\ =\dfrac{\cos \text{ }\dfrac{\pi }{2}-\cos \text{ }0}{\dfrac{\pi }{4}\sin \text{ }\dfrac{\pi }{2}-\cos 0} \\ =\dfrac{-1}{\dfrac{\pi }{4}-1} \\ =\dfrac{-1}{\text{(}\dfrac{\pi -4}{4}\text{)}} \\ =\dfrac{-4}{\pi -4} \\ =\dfrac{4}{4-\pi } [/MATH]
Not sure if my work correct or not. Can someone check my calculation? Thanks in advance.

The question asks for "equation of the tangent".

Where is the equation?

 

[mimie]

The question asks for "equation of the tangent".

Where is the equation?

I am finding the gradient of tangent before proceed to the tangent's equation part.

for equation of tangent, i will do it like this
[MATH] y-\dfrac{\pi}{4}=\dfrac{dy}{dx}(x-\dfrac{\pi}{4}) [/MATH]

 

[Dr.Peterson]

I get the same derivative, so you're good.

I did a lot less work (or at least less writing), by just taking the equation to your second line and immediately plugging in the values of the variables, and solving for y' in that form.