Mathematics - Differentiation Question with Solution | TestHub

$\left(a+\sqrt{2}b\cos x\right)\left(a-\sqrt{2}b\cos y\right)=a^2-b^2$

$\left(a-\sqrt{2}b\cos y\right)\left(-\sqrt{2}b\sin x\right)\frac{dx}{dy}+\sqrt{2}b\sin y\left(a+\sqrt{2}b\cos x\right)=0$

$\frac{dx}{dy}=\frac{\sqrt{2}b\sin y\left(a+\sqrt{2}b\cos x\right)}{\sqrt{2}b\sin x\left(a-\sqrt{2}b\cos y\right)}=\frac{\sin y\left(a+\sqrt{2}b\cos x\right)}{\sin x\left(a-\sqrt{2}b\cos y\right)}$

$\frac{dx}{dy}\left|{}_{\left(\frac{\pi }{4},\frac{\pi }{4}\right)}=\frac{a+b}{a-b}\right.$.