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Question: Find the derivative of \(\csc x\) from first principles?...

Find the derivative of csc⁡x\csc x from first principles?

Explanation

Solution

Here we have to find the derivative of the given trigonometric function. Firstly we will write the first principle formula given as f′(x)=lim⁡h→0f(x+h)−f(x)h{f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} and then we will substitute the value of function in it. Then we will simplify the value so that in don’t come in indeterminate form. Finally we will put the limit and get our desired answer.

Complete Solution
We have to find the derivative by first principle of:
csc⁡x\csc x….(1)\left( 1 \right)
Now as we know from the definition of first principle that the derivative of the function is calculated by below formula:
f′(x)=lim⁡h→0f(x+h)−f(x)h{f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}
Now from equation (1) we get f(x+h)=csc⁡(x+h)f\left( x+h \right)=\csc \left( x+h \right) substitute the values above we get,
⇒f′(x)=lim⁡h→0csc⁡(x+h)−csc⁡xh\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\csc \left( x+h \right)-\csc x}{h}…..(2)\left( 2 \right)
If we substitute the limit we will get the value as:
⇒f′(x)=lim⁡h→0csc⁡x−csc⁡x0\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\csc x-\csc x}{0}
⇒f′(x)=00\Rightarrow {f}'\left( x \right)=\dfrac{0}{0}
As we are getting indeterminate form we will simplify the value further before substituting the limit in it.
Using the relation between cosecant and sine which is csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x} in equation (2) we get,
⇒f′(x)=lim⁡h→01sin⁡(x+h)−1sin⁡xh\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{1}{\sin \left( x+h \right)}-\dfrac{1}{\sin x}}{h}
Take LCM in the numerator,
⇒f′(x)=lim⁡h→0sin⁡x−sin⁡(x+h)sin⁡(x+h)×sin⁡xh\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{\sin x-\sin \left( x+h \right)}{\sin \left( x+h \right)\times \sin x}}{h}
⇒f′(x)=lim⁡h→0sin⁡x−sin⁡(x+h)h×sin⁡(x+h)×sin⁡x\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin x-\sin \left( x+h \right)}{h\times \sin \left( x+h \right)\times \sin x}
Now use the sum relation given assin⁡A−sin⁡B=2cos⁡(A+B2)cos⁡(A−B2)\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) above where A=xA=x and B=x+hB=x+h we get,
⇒f′(x)=lim⁡h→02cos⁡(x+x+h2)cos⁡(x−x−h2)h×sin⁡(x+h)×sin⁡x\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\cos \left( \dfrac{x+x+h}{2} \right)\cos \left( \dfrac{x-x-h}{2} \right)}{h\times \sin \left( x+h \right)\times \sin x}
⇒f′(x)=lim⁡h→02cos⁡(2x+h2)cos⁡(−h2)h×sin⁡(x+h)×sin⁡x\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\cos \left( \dfrac{2x+h}{2} \right)\cos \left( \dfrac{-h}{2} \right)}{h\times \sin \left( x+h \right)\times \sin x}
Take 22 in the denominator and separate the cosine term with only hh as follows:
⇒f′(x)=lim⁡h→0cos⁡(2x+h2)sin⁡(x+h)×sin⁡xlim⁡h→0cos⁡(−h2)h2\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\cos \left( \dfrac{2x+h}{2} \right)}{\sin \left( x+h \right)\times \sin x}\displaystyle \lim_{h \to 0}\dfrac{\cos \left( \dfrac{-h}{2} \right)}{\dfrac{h}{2}}
We know cos⁡(−A)=−cos⁡A\cos \left( -A \right)=-\cos A using it above we get,
⇒f′(x)=lim⁡h→0cos⁡(2x+h2)sin⁡(x+h)×sin⁡xlim⁡h→0−cos⁡(h2)h2\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\cos \left( \dfrac{2x+h}{2} \right)}{\sin \left( x+h \right)\times \sin x}\displaystyle \lim_{h \to 0}\dfrac{-\cos \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}}
As the second limit is a standard calculus limit whose value is unity that is lim⁡h→0cosh⁡h=1\displaystyle \lim_{h \to 0}\dfrac{\cosh }{h}=1 so,
⇒f′(x)=lim⁡h→0cos⁡(2x+h2)sin⁡(x+h)×sin⁡x×−1\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\cos \left( \dfrac{2x+h}{2} \right)}{\sin \left( x+h \right)\times \sin x}\times -1
Put the limit,
⇒f′(x)=cos⁡(2x+02)sin⁡(x+0)×sin⁡x×−1\Rightarrow {f}'\left( x \right)=\dfrac{\cos \left( \dfrac{2x+0}{2} \right)}{\sin \left( x+0 \right)\times \sin x}\times -1
⇒f′(x)=cos⁡(x)sin⁡(x)×sin⁡x×−1\Rightarrow {f}'\left( x \right)=\dfrac{\cos \left( x \right)}{\sin \left( x \right)\times \sin x}\times -1
Now as cot⁡A=cos⁡xsin⁡x\cot A=\dfrac{\cos x}{\sin x} and csc⁡A=1sin⁡A\csc A=\dfrac{1}{\sin A}we get,
⇒f′(x)=cos⁡xsin⁡x×−1sin⁡x\Rightarrow {f}'\left( x \right)=\dfrac{\cos x}{\sin x}\times \dfrac{-1}{\sin x}
⇒f′(x)=−csc⁡xcot⁡x\Rightarrow {f}'\left( x \right)=-\csc x\cot x
So we got the answer as −csc⁡xcot⁡x-\csc x\cot x .

Hence derivative of csc⁡x\csc x from first principles is −csc⁡xcot⁡x-\csc x\cot x .

Note:
If f(x)f\left( x \right) is a real function in its domain and a function defined as f′(x)=lim⁡h→0f(x+h)−f(x)h{f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} exists it is known as the derivative of that function and this is our first principle. As the value was coming in indeterminate form we have simplified the value using the relation between the six trigonometric functions which are sine, cosine, tangent, secant, cosecant and cotangent.