Question

How do you differentiate $y=\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$ ?

Answer 1

We are given a sum of two different trigonometric functions to be differentiated. Therefore, we must have prior knowledge of the derivatives of sine and cosine of a function. We also know that the derivatives of sine and cosine functions are different. Since, the trigonometric functions are in the denominator, therefore they will be first raised to power of $-1$ and then differentiated.

Complete step-by-step solution:
The sine and cosine functions are written in the denominators of the terms. In order to bring them up in the numerator, we must assign them a power of $-1$.
This is because for any mathematical entity, $\dfrac{1}{x}={{x}^{-1}}$
$\Rightarrow \dfrac{1}{\sin x}+\dfrac{1}{\cos x}={{\left( \sin x \right)}^{-1}}+{{\left( \cos x \right)}^{-1}}$
Now, we shall differentiate the modified equation using the chain rule of differentiation. According to this rule, we must start differentiating with the outermost functions like the power to which a function is raised. Then, we shall proceed step-by-step to the innermost functions like the angles of any trigonometric function which can be the function the any variable.
$\dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( \sin x \right)}^{-1}}+\dfrac{d}{dx}{{\left( \cos x \right)}^{-1}}$
By the rules of basic differentiation, we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{x-1}}$. Using this for sine and cosine functions, we get
$\Rightarrow \dfrac{dy}{dx}=\left( -1 \right){{\left( \sin x \right)}^{-1-1}}.\dfrac{d}{dx}\sin x+\left( -1 \right){{\left( \cos x \right)}^{-1-1}}\dfrac{d}{dx}\cos x$
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{{{\sin }^{2}}x}.\left( \cos x \right)-\dfrac{1}{{{\cos }^{2}}x}\left( -\sin x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{\sin x}{{{\cos }^{2}}x}-\dfrac{\cos x}{{{\sin }^{2}}x} \\\ \end{aligned}$
We will now substitute the values from basic trigonometry as $\dfrac{\sin x}{\cos x}=\tan x$, $\dfrac{1}{\cos x}=\sec x$, $\dfrac{\cos x}{\sin x}=\cot x$ and $\dfrac{1}{\sin x}=\cos ecx$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\sin x}{\cos x}.\dfrac{1}{\cos x}-\dfrac{\cos x}{\sin x}.\dfrac{1}{\sin x}$
$\Rightarrow \dfrac{dy}{dx}=\tan x.\sec x-\cot x.\cos ecx$
Therefore, the derivative of $y=\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$ is $\dfrac{dy}{dx}=\tan x.\sec x-\cot x.\cos ecx$.

Note: Another method of solving this problem is by initially converting $\dfrac{1}{\sin x}$ and $\dfrac{1}{\cos x}$ into cosecant and secant functions respectively. Then, we can simply differentiate the equation as it would convert to $y=\cos ecx+\sec x$. However, in order to use this method, we must have prior knowledge of derivatives of cosecant and secant functions.