Find dydx, if y=ln(7x2+y2).

Answer 1

can also look at solving by

e^y=7x²+y²

e^yy'=14x+2yy',,,,,differentiating with respect to x where y'=dy/dx

y'=14x/(e^y-2y)

y'=14x/(7x²+y²-2y),,,,,since e^y=7x²+y² for this f(x,y)

Answer 2

Use implicit differenciation and the chain rule

y=ln(7x² + y²)

dy/dx= d/dx(ln(7x² + y²)

dy/dx = (1/(7x² + y²)) (d (7x² +y²)/dx) the bold section is being multiplied by a fraction, so it actually

becomes the numerator

dy/dx = (14x +2y dy/dx)/(7x² + y²) seperate into 2 fractions

dy/dx = 14x/(7x² + y²) + (2y/(7x² + y²) dy/dx) subtract the term containing dy/dx

dy/dx - (2y/(7x² + y²) dydx = 14x/(7x² + y²)

(1 - (2y/(7x² + y²) ) dy/dx = 14x/(7x² + y²) divide both sides by the expresson in bold