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Question: Find the derivative of the following functions from the first principle w.r.t. to x, sec 3x....

Find the derivative of the following functions from the first principle w.r.t. to x, sec 3x.

Explanation

Solution

A derivative is a measure of the rate of change. We will use the first principle, that is lim⁡h→0 f(x+h)−f(x)h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}. We take f(x) as sec 3x in the first principle to solve this question further. By simplifying we get the derivative of this function.

Complete step-by-step answer:
Let f(x) = sec 3x
Derivative of a function f(x) is given by lim⁡h→0 f(x+h)−f(x)h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}.
Applying the first principle to find the derivative of sec 3x.
lim⁡h→0 sec⁡3(x+h)−sec⁡3xh\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec 3(x+h)-\sec 3x}{h}
=lim⁡h→0 1cos⁡3(x+h)−1cos⁡3xh=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\cos 3(x+h)}-\dfrac{1}{\cos 3x}}{h}
=lim⁡h→0 cos⁡3x−cos⁡3(x+h)hcos⁡3xcos⁡3(x+h)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos 3x-\cos 3(x+h)}{h\cos 3x\cos 3(x+h)}
=lim⁡h→0 cos⁡3x−(cos⁡3xcos⁡3h−sin⁡3xsin⁡3h)hcos⁡3xcos⁡3(x+h)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos 3x-(\cos 3x\cos 3h-\sin 3x\sin 3h)}{h\cos 3x\cos 3(x+h)}
Simplifying
=lim⁡h→0 cos⁡3x(1−cos⁡3h)+sin⁡3xsin⁡3hhcos⁡3xcos⁡3(x+h)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos 3x(1-\cos 3h)+\sin 3x\sin 3h}{h\cos 3x\cos 3(x+h)}
=lim⁡h→0 cos⁡3x(1−cos⁡3h)hcos⁡3xcos⁡3(x+h)+lim⁡h→0 sin⁡3xsin⁡3hhcos⁡3xcos⁡3(x+h)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos 3x(1-\cos 3h)}{h\cos 3x\cos 3(x+h)}+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 3x\sin 3h}{h\cos 3x\cos 3(x+h)}
Simplifying
We get,
=lim⁡h→0 (1−cos⁡3h)hcos⁡3(x+h)+sin⁡3xcos⁡3x×lim⁡h→0 sin⁡3hh×lim⁡h→0 1cos⁡3(x+h)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(1-\cos 3h)}{h\cos 3(x+h)}+\dfrac{\sin 3x}{\cos 3x}\times \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 3h}{h}\times \underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos 3(x+h)}
Substituting the value of h as 0.
=0+sin⁡3xcos⁡3x×3×1cos⁡3x =3tan⁡3xsec⁡3x \begin{aligned} & =0+\dfrac{\sin 3x}{\cos 3x}\times 3\times \dfrac{1}{\cos 3x} \\\ & =3\tan 3x\sec 3x \\\ \end{aligned}.

Note: The first derivative of a function is a new function that gives the instantaneous rate of change of some desired function at any point. The first derivative of a function represents the rate of change of one variable with respect to another variable. It can be the rate of change of distance with respect to time or the temperature with respect to distance.We can use the formula,f′(x)=lim⁡h→0 f(x+h)−f(x)hf'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h} to determine an expression that describes the gradient of the graph (or the gradient of the tangent to the graph) at any point on the graph.The expression (or gradient function) is called derivative.This method is called differentiation from the first principles or using the definition.