\(\tan(5x+y)=5x\)
Differentiating both sides yields
\(\sec^2(5x+y)\times\left(5+\dfrac{\mathrm dy}{\mathrm dx}\right)=5\)
\(\dfrac{\mathrm dy}{\mathrm dx}=5\cos^2(5x+y)-5\)
At the point (0,0), you get
\(\dfrac{\mathrm dy}{\mathrm dx}=5\cos^2(5(0)+0)-5=5-5=0\)
