Derivative of tg(7x+5)*exp(3x)

1. Apply the product rule:

   $$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$$

   $f{\left(x \right)} = \tan{\left(7 x + 5 \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Rewrite the function to be differentiated:

      $$\tan{\left(7 x + 5 \right)} = \frac{\sin{\left(7 x + 5 \right)}}{\cos{\left(7 x + 5 \right)}}$$

   2. Apply the quotient rule, which is:

      $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

      $f{\left(x \right)} = \sin{\left(7 x + 5 \right)}$ and $g{\left(x \right)} = \cos{\left(7 x + 5 \right)}$.

      To find $\frac{d}{d x} f{\left(x \right)}$:

      1. Let $u = 7 x + 5$.

      2. The derivative of sine is cosine:

         $$\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$$

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(7 x + 5\right)$:

         1. Differentiate $7 x + 5$ term by term:

            1. The derivative of a constant times a function is the constant times the derivative of the function.

               1. Apply the power rule: $x$ goes to $1$

               So, the result is: $7$

            2. The derivative of the constant $5$ is zero.

            The result is: $7$

         The result of the chain rule is:

         $$7 \cos{\left(7 x + 5 \right)}$$

      To find $\frac{d}{d x} g{\left(x \right)}$:

      1. Let $u = 7 x + 5$.

      2. The derivative of cosine is negative sine:

         $$\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$$

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(7 x + 5\right)$:

         1. Differentiate $7 x + 5$ term by term:

            1. The derivative of a constant times a function is the constant times the derivative of the function.

               1. Apply the power rule: $x$ goes to $1$

               So, the result is: $7$

            2. The derivative of the constant $5$ is zero.

            The result is: $7$

         The result of the chain rule is:

         $$- 7 \sin{\left(7 x + 5 \right)}$$

      Now plug in to the quotient rule:

      $\frac{7 \sin^{2}{\left(7 x + 5 \right)} + 7 \cos^{2}{\left(7 x + 5 \right)}}{\cos^{2}{\left(7 x + 5 \right)}}$

   $g{\left(x \right)} = e^{3 x}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 3 x$.

   2. The derivative of $e^{u}$ is itself.

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 3 x$:

      1. The derivative of a constant times a function is the constant times the derivative of the function.

         1. Apply the power rule: $x$ goes to $1$

         So, the result is: $3$

      The result of the chain rule is:

      $$3 e^{3 x}$$

   The result is: $\frac{\left(7 \sin^{2}{\left(7 x + 5 \right)} + 7 \cos^{2}{\left(7 x + 5 \right)}\right) e^{3 x}}{\cos^{2}{\left(7 x + 5 \right)}} + 3 e^{3 x} \tan{\left(7 x + 5 \right)}$

2. Now simplify:

   $$7 e^{3 x} \tan^{2}{\left(7 x + 5 \right)} + 3 e^{3 x} \tan{\left(7 x + 5 \right)} + 7 e^{3 x}$$

The answer is:

$$7 e^{3 x} \tan^{2}{\left(7 x + 5 \right)} + 3 e^{3 x} \tan{\left(7 x + 5 \right)} + 7 e^{3 x}$$