Question

What is the derivative of $\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}$ ?

Answer 1

In the given problem, we are required to differentiate $\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}$ with respect to x. Since, $\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}$ is a rational trigonometric function in variable x, so we will have to apply quotient rule of differentiation in the process of differentiating $\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}$.Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly.

Complete step by step answer:
Let us assume the function to be $f(x)$. Then, we have,
$f(x) = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}$
Differentiating both sides with respect to x,
$\Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}\left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$

Now, using the quotient rule of differentiation, we know that $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$ .
So, Applying quotient rule to $\dfrac{d}{{dx}}\left( {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right)$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {\sin x - \cos x} \right)\dfrac{d}{{dx}}\left( {\sin x + \cos x} \right) - \left( {\sin x + \cos x} \right)\dfrac{d}{{dx}}\left( {\sin x - \cos x} \right)}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$
Now, separating the derivatives of the functions, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {\sin x - \cos x} \right)\left[ {\dfrac{d}{{dx}}\left( {\sin x} \right) + \dfrac{d}{{dx}}\left( {\cos x} \right)} \right] - \left( {\sin x + \cos x} \right)\left[ {\dfrac{d}{{dx}}\left( {\sin x} \right) - \dfrac{d}{{dx}}\left( {\cos x} \right)} \right]}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$

Now, we know that the derivative of $\sin x$with respect to x is $\cos x$ and $\cos x$ with respect to x is $- \sin x$. Substituting these derivatives, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {\sin x - \cos x} \right)\left[ {\cos x + \left( { - \sin x} \right)} \right] - \left( {\sin x + \cos x} \right)\left[ {\cos x - \left( { - \sin x} \right)} \right]}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$
Simplifying the expression, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {\sin x - \cos x} \right)\left[ {\cos x - \sin x} \right] - \left( {\sin x + \cos x} \right)\left[ {\cos x + \sin x} \right]}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - {{\left( {\sin x - \cos x} \right)}^2} - {{\left( {\sin x + \cos x} \right)}^2}}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$

Now, using the algebraic identities ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left[ {{{\sin }^2}x - 2\sin x\cos x + {{\cos }^2}x} \right] - \left[ {{{\sin }^2}x + 2\sin x\cos x + {{\cos }^2}x} \right]}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$
Using the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left[ {1 - 2\sin x\cos x} \right] - \left[ {1 + 2\sin x\cos x} \right]}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$
Opening the brackets and simplifying the expression,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1 + 2\sin x\cos x - 1 - 2\sin x\cos x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$
Cancelling the like terms with opposite signs, we get,
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$

Hence, the derivative of $\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}$ with respect to x is $\dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}}$.

Note: We can also further simplify the derivative expression as follows:
$\Rightarrow \dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}} = \dfrac{{ - 2}}{{{{\sin }^2}x - 2\sin x\cos x + {{\cos }^2}x}}$
The whole square term is expanded using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$,
$\Rightarrow \dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}} = \dfrac{{ - 2}}{{1 - 2\sin x\cos x}}$
Now, we use the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$. We also know the double angle formula for sine as $\sin 2x = 2\sin x\cos x$. So, we get,
$\therefore \dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}} = \dfrac{{ - 2}}{{1 - \sin 2x}} = \dfrac{2}{{\sin 2x - 1}}$
The simplified form for the derivative of the given expression is $\dfrac{2}{{\sin 2x - 1}}$.