Question

Assertion: Derivative of $3\cot x+5\text{cosec}x$ is $-\text{cosec}x\left( 3\text{cosec}x+5\cot x \right)$
Reason: ${f}'\left( a \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$
(a) Both assertion and reason are correct and Reason is correct explanation of assertion
(b) Both assertion and reason are correct but reason is not correct explanation of assertion
(c) Assertion is correct but Reason is wrong
(d) Assertion is incorrect but Reason is correct.

Answer 1

We solve this problem by using the first principle rule of differentiation that is
${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
By using this first principle rule we find the derivative of $\cot x,\text{cosec}x$ to find the derivative of $3\cot x+5\text{cosec}x$

Complete step by step answer:
We are asked to find the derivative of $3\cot x+5\text{cosec}x$
Let us assume that the given function as
$\Rightarrow f\left( x \right)=3\cot x+5\text{cosec}x$
Let us assume that the functions as

& \Rightarrow g\left( x \right)=\cot x \\\ & \Rightarrow h\left( x \right)=\text{cosec}x \\\ \end{aligned}$$ By substituting the functions in given function we get $$\Rightarrow f\left( x \right)=3g\left( x \right)+5h\left( x \right)$$ Now, by differentiating with respect to $$'x'$$ on both sides we get $$\Rightarrow {f}'\left( x \right)=3{g}'\left( x \right)+5{h}'\left( x \right).........equation(i)$$ Now, let us find the derivative of $$\cot x,\text{cosec}x$$ separately We know that the first principle of differentiation that is $${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$$ By using this principle to $$g\left( x \right)$$ we get $$\begin{aligned} & \Rightarrow {g}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{g\left( x+h \right)-g\left( x \right)}{h} \\\ & \Rightarrow {g}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\cot \left( x+h \right)-\cot x}{h} \\\ \end{aligned}$$ We know that $$\cot x=\dfrac{\cos x}{\sin x}$$ By substituting this formula in above equation we get $$\begin{aligned} & \Rightarrow {g}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{\cos \left( x+h \right)}{\sin \left( x+h \right)}-\dfrac{\cos x}{\sin x}}{h} \\\ & \Rightarrow {g}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin x.\cos \left( x+h \right)-\cos x.\sin \left( x+h \right)}{h.\sin \left( x+h \right).\sin x} \\\ \end{aligned}$$ We know that the formula of composite angle that is $$\sin \left( A+B \right)=\sin A.\cos B+\cos A.\sin B$$ By using this formula in above equation we get $$\begin{aligned} & \Rightarrow {g}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin \left( x-\left( x+h \right) \right)}{h.\sin \left( x+h \right).\sin x} \\\ & \Rightarrow {g}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin \left( h \right)}{h}.\displaystyle \lim_{h \to 0}\dfrac{-1}{\sin \left( x+h \right).\sin x} \\\ \end{aligned}$$ We know that the standard result of limits that is $$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$$ By using this result in above equation we get $$\begin{aligned} & \Rightarrow {g}'\left( x \right)=1\times \left( \dfrac{-1}{\sin x.\sin x} \right) \\\ & \Rightarrow {g}'\left( x \right)=-\text{cose}{c}^{2}x...........equation(ii) \\\ \end{aligned}$$ By using this principle to $$h\left( x \right)$$ we get $$\begin{aligned} & \Rightarrow {h}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{h\left( x+h \right)-h\left( x \right)}{h} \\\ & \Rightarrow {h}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\text{cosec}\left( x+h \right)-\text{cosec}x}{h} \\\ \end{aligned}$$ We know that $$\text{cosec}x=\dfrac{1}{\sin x}$$ By substituting this formula in above equation we get $$\begin{aligned} & \Rightarrow {h}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{1}{\sin \left( x+h \right)}-\dfrac{1}{\sin x}}{h} \\\ & \Rightarrow {h}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin x-\sin \left( x+h \right)}{h.\sin \left( x+h \right).\sin x} \\\ \end{aligned}$$ We know that the formula of half angle that is $$\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right).\cos \left( \dfrac{A+B}{2} \right)$$ By using this formula in above equation we get $$\begin{aligned} & \Rightarrow {h}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\sin \left( \dfrac{x-\left( x+h \right)}{2} \right).\cos \left( \dfrac{x+x+h}{2} \right)}{h.\sin \left( x+h \right).\sin x} \\\ & \Rightarrow {h}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin \left( \dfrac{h}{2} \right)}{h}.\displaystyle \lim_{h \to 0}\dfrac{-2\cos \left( \dfrac{2x+h}{2} \right)}{\sin \left( x+h \right).\sin x} \\\ \end{aligned}$$ We know that the standard result of limits that is $$\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{x}=a$$ By using this result in above equation we get $$\begin{aligned} & \Rightarrow {h}'\left( x \right)=\dfrac{1}{2}\times \left( \dfrac{-2\cos x}{\sin x.\sin x} \right) \\\ & \Rightarrow {h}'\left( x \right)=-\cot x\times \text{cosec}x...........equation(iii) \\\ \end{aligned}$$ Now by substituting the equation (ii) and equation (iii) in equation (i) we get $$\Rightarrow {f}'\left( x \right)=3\left( -\text{cose}{c}^{2}x \right)+5\left( -\cot x.\text{cosec}x \right)$$ Now by taking the common term out we get $$\Rightarrow {f}'\left( x \right)=-\text{cosec}x\left( 3\text{cosec}x+5\cot x \right)$$ Therefore the derivative of $$3\cot x+5\text{cosec}x$$ is $$-\text{cosec}x\left( 3\text{cosec}x+5\cot x \right)$$ Here, we can see that the assertion is correct and reason is correct and Reason is the correct explanation of assertion because we used the formula in reason to prove assertion. **So, the correct answer is “Option a”.** **Note:** Students may make mistakes in solving the problem. We know that the standard derivative formulas that is $$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( \cot x \right)=-\text{cose}{c}^{2}x \\\ & \Rightarrow \dfrac{d}{dx}\left( \text{cosec}x \right)=-\cot x.\text{cosec}x \\\ \end{aligned}$$ We can use these formulas directly. But we somehow need to prove that the reason mentioned in the question is linked to assertion. This means that we need to use the reason to prove the assertion in these types of questions. So, we need to solve the problem using the first principle that is $${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$$ This is the correct procedure.