Solve by using first principle

A) f(x) = 3x² + x
B) f(x) = tanx

A) To find the derivative of f(x) = 3x² + x using the first principle, we will start by evaluating the difference quotient.

The difference quotient is given by:

[f(x + h) - f(x)] / h

= [(3(x + h)² + (x + h)) - (3x² + x)] / h

= [(3(x² + 2hx + h²) + x + h) - (3x² + x)] / h

= [(3x² + 6hx + 3h² + x + h) - (3x² + x)] / h

= (3h² + 6hx + h) / h

Cancel out the common factor of h:

= 3h + 6x + 1

Now, taking the limit as h approaches 0:

lim(h→0) (3h + 6x + 1) = 6x + 1

Therefore, the derivative of f(x) = 3x² + x is 6x + 1.

B) To find the derivative of f(x) = tanx using the first principle, we will again evaluate the difference quotient.

The difference quotient is given by:

[f(x + h) - f(x)] / h

= [tan(x + h) - tanx] / h

Since tan(x + h) cannot be simplified easily, we will use the trigonometric identity:

tan(x + h) - tanx = [tan(x) + tan(h)] / [1 - tan(x)tan(h)]

Therefore, the difference quotient becomes:

= [tan(x) + tan(h)] / [h * (1 - tan(x)tan(h))]

Now, taking the limit as h approaches 0:

lim(h→0) [tan(x) + tan(h)] / [h * (1 - tan(x)tan(h))]

Using the properties of limits and trigonometric identities, we can simplify this expression further:

= sec²(x) / (1 - tan²(x))

= sec²(x) / sec²(x) (since 1 - tan²(x) = sec²(x))

= 1

Therefore, the derivative of f(x) = tanx is 1.

A) To find the derivative of f(x) = 3x² + x using the first principle, we need to evaluate the following limit:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

Let's begin by calculating f(x + h):

f(x + h) = 3(x + h)² + (x + h)
= 3(x² + 2hx + h²) + x + h
= 3x² + 6hx + 3h² + x + h

Now, we substitute f(x) and f(x + h) into the formula:

f'(x) = lim(h->0) [3x² + 6hx + 3h² + x + h - (3x² + x)] / h
= lim(h->0) [6hx + 3h² + h] / h

Expanding further:

f'(x) = lim(h->0) 6hx/h + 3h²/h + h/h
= lim(h->0) 6x + 3h + 1

Since h approaches zero, we can eliminate the h-terms:

f'(x) = 6x + 1

Therefore, the derivative of f(x) = 3x² + x using the first principle is f'(x) = 6x + 1.

B) To find the derivative of f(x) = tan(x) using the first principle, we need to evaluate the following limit:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

Let's begin by calculating f(x + h):

f(x + h) = tan(x + h)

Now, we substitute f(x) and f(x + h) into the formula:

f'(x) = lim(h->0) [tan(x + h) - tan(x)] / h

By using the trigonometric identity for tan(x + h) - tan(x), we have:

f'(x) = lim(h->0) [tan(x) + tan(h) - tan(x)] / h
= lim(h->0) [tan(h)] / h

This limit does not have a finite value, so we cannot find the derivative of f(x) = tan(x) using the first principle. The derivative of tan(x) can be found using other differentiation rules or formulas, such as the chain rule.