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

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

Explanation

Solution

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)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
here f(x)=tan⁡(x)f(x) = \tan (x) and f(x+h)=tan⁡(x+h)f(x + h) = \tan (x + h) and we know that the tangent trigonometry ratio can be written as tan⁡(x)=sin⁡(x)cos⁡(x)\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}}, therefore first we simplify the f(x+h)−f(x)f(x + h) - f(x). So we have
⇒f(x+h)−f(x)=tan⁡(x+h)−tan⁡(x)\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.
⇒f(x+h)−f(x)=sin⁡(x+h)cos⁡(x+h)−sin⁡(x)cos⁡(x)\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⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B
⇒f(x+h)−f(x)=sin⁡(x).cos⁡(h)+sin⁡(h).cos⁡(x)cos⁡(x).cos⁡(h)−sin⁡(x).sin⁡(h)−sin⁡(x)cos⁡(x)\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
⇒f(x+h)−f(x)=cos⁡(x)(sin⁡(x).cos⁡(h)+sin⁡(h).cos⁡(x))−(cos⁡(x).cos⁡(h)−sin⁡(x).sin⁡(h))sin⁡(x)cos⁡(x)(cos⁡(x).cos⁡(h)−sin⁡(x).sin⁡(h))\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)}}
⇒f(x+h)−f(x)=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 (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)}}
⇒f(x+h)−f(x)=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{{{{\cos }^2}(x)\sin (h) + {{\sin }^2}(x)\sin (h)}}{{{{\cos }^2}(x).\cos (h) - \cos (x)\sin (x).\sin (h)}}
⇒f(x+h)−f(x)=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) + {{\sin }^2}(x))}}{{{{\cos }^2}(x).\cos (h) - \cos (x)\sin (x).\sin (h)}}
⇒f(x+h)−f(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).\cos (h) - \cos (x)\sin (x).\sin (h)}}
Divide by cos⁡2(x)cos⁡(h){\cos ^2}(x)\cos (h)
⇒f(x+h)−f(x)=sec⁡2xtan⁡(h)1−tan⁡(x).tan⁡(h)\Rightarrow f(x + h) - f(x) = \dfrac{{{{\sec }^2}x\tan (h)}}{{1 - \tan (x).\tan (h)}}
Now divide by h we have
⇒f(x+h)−f(x)h=1h×sec⁡2xtan⁡(h)1−tan⁡(x).tan⁡(h)\Rightarrow \dfrac{{f(x + h) - f(x)}}{h} = \dfrac{1}{h} \times \dfrac{{{{\sec }^2}x\tan (h)}}{{1 - \tan (x).\tan (h)}}
⇒f(x+h)−f(x)h=sec⁡2x×tan⁡(h)h×11−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
⇒lim⁡h→0f(x+h)−f(x)h=lim⁡h→0(sec⁡2x×tan⁡(h)h×11−tan⁡(x).tan⁡(h))\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)
⇒lim⁡h→0f(x+h)−f(x)h=lim⁡h→0sec⁡2x×lim⁡h→0tan⁡(h)h×lim⁡h→011−tan⁡(x).tan⁡(h)\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 lim⁡h→0tan⁡(h)h=1\mathop {\lim }\limits_{h \to 0} \dfrac{{\tan (h)}}{h} = 1 and lim⁡h→0tan⁡(h)=0\mathop {\lim }\limits_{h \to 0} \tan (h) = 0substituting these we get
⇒lim⁡h→0f(x+h)−f(x)h=sec⁡2x×1×lim⁡h→011−0\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}}
⇒lim⁡h→0f(x+h)−f(x)h=sec⁡2x\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} = {\sec ^2}x
⇒f′(x)=sec⁡2x\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)=lim⁡h→0=f(x+h)−f(x)hf'(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