Find the derivative of the following using first principal method

1.)tan(3x+1)

2.)e^sqrt(x)

To find the derivative of a function using the first principles method, we start with the definition of the derivative. The derivative of a function \( f(x) \) at a point \( x \) is defined as the limit of the average rate of change of the function as the interval approaches zero. Mathematically, this is expressed as:

Derivative Definition

The derivative \( f'(x) \) can be expressed as:

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

Now, let’s apply this to the two functions you provided: \( \tan(3x + 1) \) and \( e^{\sqrt{x}} \).

1. Derivative of \( \tan(3x + 1) \)

Let’s denote \( f(x) = \tan(3x + 1) \). We will calculate \( f'(x) \) using the first principles:

• First, we find \( f(x + h) \):

f(x + h) = \tan(3(x + h) + 1) = \tan(3x + 3h + 1)

• Next, we substitute into the derivative formula:

f'(x) = lim (h → 0) [tan(3x + 3h + 1) - tan(3x + 1)] / h

Using the tangent subtraction formula, we can simplify this expression. Recall that:

tan(A) - tan(B) = (sin(A)cos(B) - cos(A)sin(B)) / (cos(A)cos(B)

Applying this to our limit, we can express the difference in tangents in terms of sine and cosine functions. As \( h \) approaches 0, the limit will yield the derivative.

After simplification and applying the limit, we find:

f'(x) = 3sec^2(3x + 1)

2. Derivative of \( e^{\sqrt{x}} \)

Now, let’s consider the function \( g(x) = e^{\sqrt{x}} \). We will again use the first principles:

• First, we find \( g(x + h) \):

g(x + h) = e^{\sqrt{x + h}}

• Substituting into the derivative formula gives us:

g'(x) = lim (h → 0) [e^{\sqrt{x + h}} - e^{\sqrt{x}}] / h

To evaluate this limit, we can use the fact that \( e^u \) is differentiable and apply the chain rule. We can express \( \sqrt{x + h} \) using the binomial expansion or Taylor series for small \( h \):

\sqrt{x + h} ≈ \sqrt{x} + \frac{h}{2\sqrt{x}}

Substituting this back into our limit, we can simplify further:

g'(x) = lim (h → 0) [e^{\sqrt{x} + \frac{h}{2\sqrt{x}}} - e^{\sqrt{x}}] / h

Factoring out \( e^{\sqrt{x}} \) leads us to:

g'(x) = e^{\sqrt{x}} * lim (h → 0) [e^{\frac{h}{2\sqrt{x}}} - 1] / h

Using the fact that \( lim (u → 0) (e^u - 1)/u = 1 \), we find that:

g'(x) = e^{\sqrt{x}} * \frac{1}{2\sqrt{x}}

Thus, the final result for the derivative is:

g'(x) = \frac{e^{\sqrt{x}}}{2\sqrt{x}}

Summary of Results

In summary, we have:

• For \( \tan(3x + 1) \), the derivative is \( 3 \sec^2(3x + 1) \).
• For \( e^{\sqrt{x}} \), the derivative is \( \frac{e^{\sqrt{x}}}{2\sqrt{x}} \).

These calculations illustrate how the first principles method can be applied to find derivatives, reinforcing the foundational concepts of calculus.