How To Calculate Dy/dx

How do I calculate dy/dx of y=(x-1) /(x+1)?

We use the u/v formula which states thatd(u/v)/dx= (vdu/dx-udv/dx)/v^2so dy/dx if y=(x-1)/(x+1) is 2/(x+1)^2

How do I calculate dy/dx if Ye^x +Xe^y + X + Y =0?

y e^x + xe^y + x + y =0to start, remember d[f(x)f(y)] = f(y) d(f(x) dx + f(x) d(f(y) dye^x dy +y e^x dx + e^y dx + x e^y dy + dx + dy =0[e^x dy +x e^y dy +dy] + [y e^x dx +e^y dx+ dx] = 0dy[ e^x +x e^y +1] + dx [y e^x +e^y + 1] = 0dy[ e^x +x e^y +1] = - dx [y e^x +e^y + 1]dy/dx = - [y e^x +e^y + 1]/[ e^x +x e^y +1]

How do you calculate dx/dy as opposed to dy/dx?

dy/dx means you differentiate y with respect to x, or differentiate implicitly and then divide by dx; So to calculate dx/dy, differentiate x with respect to y, or differentiate implicitly and then divide by dy.Or if you've already calculated dy/dx, then simply take it's reciprocal as dx/dy.For example, from the equation ax^n + by^m = c, let's find:(i) dy/dx;(ii) dx/dy.Differentiating ax^n + by^m = c, we getanx^(n-1)dx + bmy^(m-1)dy = 0So,(i) to find dy/dx we divide by dx to getanx^(n-1) + bmy^(m-1)(dy/dx) = 0Regrouping,bmy^(m-1)(dy/dx) = anx^(n-1)Then we havedy/dx = [anx^(n-1)]/[bmy^(m-1)].(ii) by taking the reciprocal of dy/dx above, we getdx/dy = [bmy^(m-1)]/[anx^(n-1)].

How do you calculate [math]dy/dx=2ax+b[/math] from [math]y=ax^2+bx+c[/math]?

Since [math]y[/math] is a function of [math]x[/math], we can write [math]y = f(x) [/math]We have [math]f(x) = ax^2 + bx + c[/math] and we have to compute the derivative of this function w.r.t [math]x[/math] which we denote as: [math]\dfrac{dy}{dx}[/math] or [math]f’(x).[/math]Recall the definition of derivative (given that the function is differentiable in the specified domain):[math]\displaystyle f’(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}[/math][math]\implies \displaystyle f’(x) = \lim_{h \to 0} \dfrac{a(x+h)^2+b(x+h)+c - ax^2-bx-c}{h}[/math][math]\implies \displaystyle f’(x) = \lim_{h \to 0} \dfrac{a(x^2+h^2+2hx)+b(x+h)- ax^2-bx}{h}[/math][math]\implies \displaystyle f’(x) = \lim_{h \to 0} \dfrac{ax^2+ah^2+2ahx+bx+bh- ax^2-bx}{h}[/math][math]\implies \displaystyle f’(x) = \lim_{h \to 0} \dfrac{ah^2+2ahx+bh}{h}[/math][math]\implies \displaystyle f’(x) = \lim_{h \to 0} \dfrac{h(ah+2ax+b)}{h}[/math][math]\implies \displaystyle f’(x) = \lim_{h \to 0} \ ah+2ax+b = \boxed{2ax+b}[/math]In this case, we had a Parabola [math]y = ax^2 + bx + c[/math] which we know is differentiable on its entire domain.However, to check if a function is differentiable or not at a particular point [math]x =a, [/math]we compute the Left Hand Derivative([math]L.H.D[/math]) and the Right Hand Derivate([math]R.H.D[/math]). If they’re equal and finite, the function is differentiable at [math]x=a.[/math]