Question

How do you find the derivative of \[y = \tan (x)\] using first principles?

Answer 1

Hint: Here in this question, we consider the given function as y and we are going to differentiate the given function with respect to x. The function is a trigonometric function so to differentiate the function we use the standard formulas on the trigonometry and then we are going to simplify.

Complete step by step solution:
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
Now here in this question we have to find the derivative of a given function by using the first principle. Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to \[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
here \[f(x) = \tan (x)\] and \[f(x + h) = \tan (x + h)\] and we know that the tangent trigonometry ratio can be written as \[\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}}\], therefore first we simplify the \[f(x + h) - f(x)\]. So we have
\[ \Rightarrow f(x + h) - f(x) = \tan (x + h) - \tan (x)\]
The tangent trigonometry ratio is written in the form of sine and cosine trigonometry ratio.
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{\sin (x + h)}}{{\cos (x + h)}} - \dfrac{{\sin (x)}}{{\cos (x)}}\]
Use the trigonometry formulas and it is defined as \[\sin (A + B) = \sin A\cos B + \cos A\sin B\] and \[\cos (A + B) = \cos A\cos B - \sin A\sin B\]
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{\sin (x).\cos (h) + \sin (h).\cos (x)}}{{\cos (x).\cos (h) - \sin (x).\sin (h)}} - \dfrac{{\sin (x)}}{{\cos (x)}}\]
On taking the LCM we get
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{\cos (x)\left( {\sin (x).\cos (h) + \sin (h).\cos (x)} \right) - \left( {\cos (x).\cos (h) - \sin (x).\sin (h)} \right)\sin (x)}}{{\cos (x)\left( {\cos (x).\cos (h) - \sin (x).\sin (h)} \right)}}\]
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{\cos (x)\sin (x)\cos (h) + {{\cos }^2}(x)\sin (h) - \cos (x).\cos (h)\sin (x) + {{\sin }^2}(x)\sin (h)}}{{{{\cos }^2}(x).\cos (h) - \cos (x)\sin (x).\sin (h)}}\]
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{{{\cos }^2}(x)\sin (h) + {{\sin }^2}(x)\sin (h)}}{{{{\cos }^2}(x).\cos (h) - \cos (x)\sin (x).\sin (h)}}\]
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{\sin (h)({{\cos }^2}(x) + {{\sin }^2}(x))}}{{{{\cos }^2}(x).\cos (h) - \cos (x)\sin (x).\sin (h)}}\]
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{\sin (h)}}{{{{\cos }^2}(x).\cos (h) - \cos (x)\sin (x).\sin (h)}}\]
Divide by \[{\cos ^2}(x)\cos (h)\]
\[ \Rightarrow f(x + h) - f(x) = \dfrac{{{{\sec }^2}x\tan (h)}}{{1 - \tan (x).\tan (h)}}\]
Now divide by h we have
\[ \Rightarrow \dfrac{{f(x + h) - f(x)}}{h} = \dfrac{1}{h} \times \dfrac{{{{\sec }^2}x\tan (h)}}{{1 - \tan (x).\tan (h)}}\]
\[ \Rightarrow \dfrac{{f(x + h) - f(x)}}{h} = {\sec ^2}x \times \dfrac{{\tan (h)}}{h} \times \dfrac{1}{{1 - \tan (x).\tan (h)}}\]
On applying the limit we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {{{\sec }^2}x \times \dfrac{{\tan (h)}}{h} \times \dfrac{1}{{1 - \tan (x).\tan (h)}}} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} = \mathop {\lim }\limits_{h \to 0} {\sec ^2}x \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan (h)}}{h} \times \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{1 - \tan (x).\tan (h)}}\]
We know that \[\mathop {\lim }\limits_{h \to 0} \dfrac{{\tan (h)}}{h} = 1\] and \[\mathop {\lim }\limits_{h \to 0} \tan (h) = 0\]substituting these we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} = {\sec ^2}x \times 1 \times \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{1 - 0}}\]
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} = {\sec ^2}x\]
\[ \Rightarrow f'(x) = {\sec ^2}x\]

Note: The differentiation is defined as the derivative of a function with respect to the independent variable. Here the dependent variable is y and the independent variable is x. If the function is to differentiate by using the first principle we use the formula and it is defined as \[f'(x) = \mathop {\lim }\limits_{h \to 0} = \dfrac{{f(x + h) - f(x)}}{h}\] , By using the limit and trigonometry formulas we can obtain the result