What is the derivative of #(sinx + cosx) / (sinx - cosx)#?

1 Answer
Jul 28, 2015

Answer:

#y^' = -2/(sinx - cosx)^2#

Explanation:

Start by taking a look at your function

#y = (sinx + cosx)/(sinx - cosx)#

Notice that this function is actually the quotient of two other functions, let's call them #f(x)# and #g(x)#

#{(f(x) = sinx + cosx), (g(x) = sinx - cosx) :}#

This means that you can differentiate this function by using the quotient rule, which allows you to find the derivative of a function that's the quotient of two other functions by using the formula

#color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2#, where #g(x) !=0#

You also need to remember that

#d/dx(sinx) = cosx#

and that

#d/dx(cosx) = -sinx#

So, calculate the derivative of #f(x)#

#d/dx(f(x)) = d/dx(sinx + cosx)#

#d/dx(f(x)) = d/dx(sinx) + d/dx(cosx)#

#d/dx(f(x)) = cosx - sinx#

and that of #g(x)#

#d/dx(g(x)) = d/dx(sinx - cosx#

#d/dx(g(x)) = d/dx(sinx) - d/dx(cosx)#

#d/dx(g(x)) = cosx - (-sinx) = cosx + sinx#

The derivative of #y# will thus look like this

#d/dx(y) = ((cosx - sinx) * (sinx - cosx) - (sinx + cosx) * (cosx + sinx))/(sinx - cosx)^2#

I'll break this into two fractions to make the calculations easier to read

#d/dx(y) = y^' = underbrace((f^'(x) * g(x))/[g(x)^2])_(color(red)(A)) - underbrace((f(x) * g^'(x))/[g(x)^2])_(color(red)(B)) = color(red)(A) - color(red)(B)#

The first fraction will be equal to

#color(red)(A) = ((cosx - sinx) * (sinx - cosx))/(sinx - cosx)^2#

#color(red)(A) = (cosx * sinx - cos^2x - sin^2x + sinx cosx)/(sinx - cosx)^2#

#color(red)(A) = (2 sinx cosx - (sin^2x + cos^2x))/(sinx - cosx)^2#

You can further simplify this by using the fact that

#color(blue)(sin^2x + cos^2x = 1)#

#color(red)(A) = (2 sinx cosx - 1)/(sinx - cosx)^2#

The second fraction will be equal to

#color(red)(B) = ((sinx + cosx) * (cosx + sinx))/(sinx - cosx)^2#

#color(red)(B) = (sinx + cosx)^2/(sinx - cosx)^2#

You can simplify this by using

#color(blue)((a + b)^2 = a^2 + 2ab + b^2)#

to get

#color(red)(B) = (sin^2x + 2 sinx cosx + cos^2x)/(sinx - cosx)^2#

#color(red)(B) = (2 sinx cosx + 1)/(sinx - cosx)^2#

Put the two fractions back together to get

#y^' = (2 sinx cosx -1 - (2 sinx cosx + 1))/(sinx - cosx)^2#

#y^' = (color(red)(cancel(color(black)(2sinxcosx))) - 1 - color(red)(cancel(color(black)(2sinxcosx))) - 1)/(sinx - cosx)^2#

Finally, you get

#y^' = color(green)(-2/(sinx - cosx)^2)#