Question

Find the derivative of \[\cos x\] from the first principle.

Answer 1

Hint: Use the first principle for finding the first derivative of any function which says that first derivative of any function \[f\left( x \right)\] can be found by using the formula \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\] and simplifying it to get the derivative.

Complete step-by-step answer:

We have the function \[y=\cos x\]. We have to find its first derivative using the first principle.
First principle says that first derivative of the function \[y=f\left( x \right)\] can be found out by evaluating the value of \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\].

Substituting \[f\left( x \right)=\cos x\], we have the first derivative of the function as \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( x+h \right)-\cos \left( x \right)}{h}\].
We know that \[\cos a-\cos b=2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{b-a}{2} \right)\].

Substituting \[a=x+h,b=x\] in the above equation, we have \[\cos \left( x+h \right)-\cos \left( x \right)=2\sin \left( \dfrac{x+h+x}{2} \right)\sin \left( \dfrac{x-\left( x+h \right)}{2} \right)\].

Thus, we have \[\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( x+h \right)-\cos \left( x \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sin \left( \dfrac{x+h+x}{2} \right)\sin \left( \dfrac{x-\left( x+h \right)}{2} \right)}{h}\].

Simplifying the above expression, we have \[\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sin \left( \dfrac{x+h+x}{2} \right)\sin \left( \dfrac{x-\left( x+h \right)}{2} \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+\dfrac{h}{2} \right)\sin \left( \dfrac{-h}{2} \right)}{\dfrac{h}{2}}\].

We know that \[\sin \left( -\theta \right)=-\sin \theta \].

Thus, we have \[\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+\dfrac{h}{2} \right)\sin \left( \dfrac{-h}{2} \right)}{\dfrac{h}{2}}=-\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+\dfrac{h}{2} \right)\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}}\].

We know that \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{x}{2} \right)}{\dfrac{x}{2}}=1\].

Thus, we have \[\dfrac{dy}{dx}=-\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+\dfrac{h}{2} \right)\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}}=-\underset{h\to 0}{\mathop{\lim }}\,\sin \left( x+\dfrac{h}{2} \right)\times 1\].

Applying the limit, we have \[\dfrac{dy}{dx}=-\underset{h\to 0}{\mathop{\lim }}\,\sin \left( x+\dfrac{h}{2} \right)\times 1=-\sin \left( x+0 \right)=-\sin x\].

Hence, the first derivative of the function \[y=\cos x\] is \[\dfrac{dy}{dx}=-\sin x\].

Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1).

We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

Note: The instantaneous rate of change of a function with respect to the dependent variable is called the derivative of the function. The first derivative of any function represents the slope of the curve of that function. One must be careful while using limits to find the value of derivatives. Directly applying the limits will give an incorrect answer.