Question

How do you find the derivative of $\sin x\left( \tan x \right)$ ? \[\]

Answer 1

Hint: We recall the product rule of differentiation where the differentiation of product of two functions $u\left( x \right)\times v\left( x \right)$ is given by $\dfrac{d}{dx}\left( u\left( x \right)\cdot v\left( x \right) \right)=u\left( x \right)\dfrac{d}{dx}v\left( x \right)+u\left( x \right)\dfrac{d}{dx}v\left( x \right)$. We take the given function as a product of $u\left( x \right)=\sin x,v\left( x \right)=\tan x$ and use the product rule. We simplify the differentiation. \[\]

Complete step by step answer:
We know from calculus that the derivative of a function of a real variable measures the rate of change of the functional value with respect to argument or input value. The process of finding derivative is called differentiation. If $f\left( x \right)$ is real valued function then we use the differential operator $\dfrac{d}{dx}$ and find the derivative as
\[\dfrac{d}{dx}f\left( x \right)={{f}^{'}}\left( x \right)\]
We know the rule to differentiate trigonometric functions.
\[\begin{align}
  & \dfrac{d}{dx}\left( \sin x \right)=\cos x,\dfrac{d}{dx}\left( \cos x \right)=-\sin x \\
 & \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x,\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x \\
 & \dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x,\dfrac{d}{dx}\left( \csc x \right)=-\csc x\cot x \\
\end{align}\]
We know product rule is used to differentiate when two or more functions multiplied with each other. If $u\left( x \right),v\left( x \right)$ are real valued differentiable functions then we can differentiate their product $u\left( x \right)v\left( x \right)$ using the following rule
\[\begin{align}
  & \dfrac{d}{dx}\left( u\left( x \right)\cdot v\left( x \right) \right)=u\left( x \right)\dfrac{d}{dx}v\left( x \right)+v\left( x \right)\dfrac{d}{dx}u\left( x \right) \\
 & \Rightarrow {{\left( uv \right)}^{'}}=u{{v}^{'}}+v{{u}^{'}} \\
\end{align}\]
We are asked in the question to differentiate the following function:$\sin x\left( \tan x \right)$. We see that we are given two functions $\sin x,\tan x$ which are multiplied to each other. We take $u\left( x \right)=\sin x,v\left( x \right)=\tan x$ and differentiate using product rule to have;
\[\dfrac{d}{dx}\left( \sin x\cdot \left( \tan x \right) \right)=\sin x\dfrac{d}{dx}\tan x+\tan x\dfrac{d}{dx}\sin x\]
We use differentiation of tangent trigonometric function $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ and sine function $\dfrac{d}{dx}\left( \sin x \right)=\cos x$ in the above step to have
\[\dfrac{d}{dx}\left( \sin x\cdot \left( \tan x \right) \right)=\sin x\cdot {{\sec }^{2}}x+\tan x\cdot \cos x\]
We convert the tangent and secant function into sine and cosine in the above step using the rule $\sec \theta =\dfrac{1}{\cos \theta },\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to have;
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \sin x\cdot \left( \tan x \right) \right)=\sin x\cdot \dfrac{1}{{{\cos }^{2}}x}+\dfrac{\sin x}{\cos x}\cdot \cos x \\
 & \Rightarrow \dfrac{d}{dx}\left( \sin x\cdot \left( \tan x \right) \right)=\dfrac{\sin x}{\cos x}\cdot \dfrac{1}{\cos x}+\sin x \\
 & \Rightarrow \dfrac{d}{dx}\left( \sin x\cdot \left( \tan x \right) \right)=\tan x\cdot \sec x+\sin x \\
\end{align}\]

Note: We note that if one of $u\left( x \right),v\left( x \right)$is not differentiable in the real number set then the product rule cannot be applied since $u\left( x \right)v\left( x \right)$ becomes non-differentiable. The sum rule states if two real functions are added to each other then their differentiation is the sum differentiations of each function which means ${{\left( u+v \right)}^{'}}={{u}^{'}}+{{v}^{'}}$ and similarly quotient rule is given by ${{\left( \dfrac{u}{v} \right)}^{'}}=\dfrac{v{{u}^{'}}-u{{v}^{'}}}{{{v}^{2}}}$.