Derivative of 3cos6t-4tg5t

1. Differentiate $3 \cos{\left(6 t \right)} - 4 \tan{\left(5 t \right)}$ term by term:

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

      1. Let $u = 6 t$.

      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 $\frac{d}{d t} 6 t$:

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

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

            So, the result is: $6$

         The result of the chain rule is:

         $$- 6 \sin{\left(6 t \right)}$$

      So, the result is: $- 18 \sin{\left(6 t \right)}$

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

      1. Rewrite the function to be differentiated:

         $$\tan{\left(5 t \right)} = \frac{\sin{\left(5 t \right)}}{\cos{\left(5 t \right)}}$$

      2. Apply the quotient rule, which is:

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

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

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

         1. Let $u = 5 t$.

         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 $\frac{d}{d t} 5 t$:

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

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

               So, the result is: $5$

            The result of the chain rule is:

            $$5 \cos{\left(5 t \right)}$$

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

         1. Let $u = 5 t$.

         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 $\frac{d}{d t} 5 t$:

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

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

               So, the result is: $5$

            The result of the chain rule is:

            $$- 5 \sin{\left(5 t \right)}$$

         Now plug in to the quotient rule:

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

      So, the result is: $- \frac{4 \left(5 \sin^{2}{\left(5 t \right)} + 5 \cos^{2}{\left(5 t \right)}\right)}{\cos^{2}{\left(5 t \right)}}$

   The result is: $- \frac{4 \left(5 \sin^{2}{\left(5 t \right)} + 5 \cos^{2}{\left(5 t \right)}\right)}{\cos^{2}{\left(5 t \right)}} - 18 \sin{\left(6 t \right)}$

2. Now simplify:

   $$\frac{2 \left(\frac{9 \sin{\left(4 t \right)}}{4} - \frac{9 \sin{\left(6 t \right)}}{2} - \frac{9 \sin{\left(16 t \right)}}{4} - 10\right)}{\cos^{2}{\left(5 t \right)}}$$

The answer is:

$$\frac{2 \left(\frac{9 \sin{\left(4 t \right)}}{4} - \frac{9 \sin{\left(6 t \right)}}{2} - \frac{9 \sin{\left(16 t \right)}}{4} - 10\right)}{\cos^{2}{\left(5 t \right)}}$$