If y=¡Ìsecx-tanx¡¡Ìsecx+tanx,show that dy¡dx=secx(tanx+secx)
Not sure what all that introductory stuff is, but if
y = secx + tanx
dy/dx = secx tanx + sec^2x
= secx(tanx+secx)
To show that dy/dx = secx(tanx + secx), we need to differentiate the given expression y = √(secx - tanx) / √(secx + tanx) with respect to x.
To begin, let's simplify the expression by rationalizing the numerator and the denominator. Recall that multiplying a number by its conjugate eliminates the square root in the denominator. Applying this concept, we can multiply both the numerator and denominator by √(secx - tanx) to rationalize the denominator:
y = √(secx - tanx) / √(secx + tanx)
Multiply the numerator and denominator by √(secx - tanx):
y = (√(secx - tanx) * √(secx - tanx)) / (√(secx + tanx) * √(secx - tanx))
Simplifying the denominator by distributing the multiplication:
y = (√[(secx - tanx)^2]) / (√(secx + tanx) * √(secx - tanx))
Simplifying the numerator by squaring the expression inside the square root:
y = (secx - tanx) / (√(secx + tanx) * √(secx - tanx))
Now, we can differentiate both sides of the equation with respect to x.
Starting with the left side, we differentiate the expression y with respect to x using the quotient rule:
dy/dx = [(secx + tanx) * d/dx(secx - tanx) - (secx - tanx) * d/dx(secx + tanx)] / (secx + tanx)^2
Next, we differentiate the terms in the numerator:
d/dx(secx - tanx) = secx * secx - secx * tanx
= sec^2(x) - sec(x) * tan(x)
d/dx(secx + tanx) = secx * secx + secx * tanx
= sec^2(x) + sec(x) * tan(x)
Plugging the values back into the differentiation formula:
dy/dx = [(secx + tanx) * (sec^2(x) - sec(x) * tan(x)) - (secx - tanx) * (sec^2(x) + sec(x) * tan(x))] / (secx + tanx)^2
Simplifying further, we have:
dy/dx = [(sec^3(x) - sec(x) * tan^2(x) + (sec^3(x) + sec(x) * tan^2(x))] / (secx + tanx)^2
Combining like terms in the numerator:
dy/dx = 2 * sec^3(x) / (secx + tanx)^2
Now, let's simplify the expression further:
dy/dx = sec^2(x) * sec(x) / (secx + tanx)^2
Finally, using the identity sec^2(x) = (secx + tanx) * sec(x), we can substitute it into the expression:
dy/dx = sec(x) * (secx + tanx) / (secx + tanx)^2
Canceling out the common factor (secx + tanx) in the numerator and denominator:
dy/dx = sec(x) / (secx + tanx)
Therefore, we have shown that dy/dx = secx(tanx + secx).