F(x)= 8sec²xtan³x + 16sec⁴xtanx, then f'(x)=?

To differentiate the function \( F(x) = 8\sec^2(x)\tan^3(x) + 16\sec^4(x)\tan(x) \), we will apply the product rule, chain rule, and the known derivatives of trigonometric functions.

### Step 1: Differentiate \( 8\sec^2(x)\tan^3(x) \)
This is a product of two functions: \( u(x) = 8\sec^2(x) \) and \( v(x) = \tan^3(x) \).

Using the product rule:
\[
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
\]

First, differentiate \( u(x) = 8\sec^2(x) \):
\[
u'(x) = 8 \cdot 2\sec^2(x)\tan(x) = 16\sec^2(x)\tan(x)
\]

Next, differentiate \( v(x) = \tan^3(x) \) using the chain rule:
\[
v'(x) = 3\tan^2(x) \cdot \sec^2(x)
\]

Now apply the product rule:
\[
\frac{d}{dx}\left(8\sec^2(x)\tan^3(x)\right) = 16\sec^2(x)\tan(x)\tan^3(x) + 8\sec^2(x) \cdot 3\tan^2(x)\sec^2(x)
\]
\[
= 16\sec^2(x)\tan^4(x) + 24\sec^4(x)\tan^2(x)
\]

### Step 2: Differentiate \( 16\sec^4(x)\tan(x) \)
This is again a product of two functions: \( u(x) = 16\sec^4(x) \) and \( v(x) = \tan(x) \).

Using the product rule again:
\[
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
\]

First, differentiate \( u(x) = 16\sec^4(x) \):
\[
u'(x) = 16 \cdot 4\sec^4(x)\tan(x) = 64\sec^4(x)\tan(x)
\]

Next, differentiate \( v(x) = \tan(x) \):
\[
v'(x) = \sec^2(x)
\]

Now apply the product rule:
\[
\frac{d}{dx}\left(16\sec^4(x)\tan(x)\right) = 64\sec^4(x)\tan(x)\tan(x) + 16\sec^4(x)\sec^2(x)
\]
\[
= 64\sec^4(x)\tan^2(x) + 16\sec^6(x)
\]

### Step 3: Combine the results
Now sum up the two parts:
\[
f'(x) = \left(16\sec^2(x)\tan^4(x) + 24\sec^4(x)\tan^2(x)\right) + \left(64\sec^4(x)\tan^2(x) + 16\sec^6(x)\right)
\]

Simplify:
\[
f'(x) = 16\sec^2(x)\tan^4(x) + (24\sec^4(x)\tan^2(x) + 64\sec^4(x)\tan^2(x)) + 16\sec^6(x)
\]
\[
f'(x) = 16\sec^2(x)\tan^4(x) + 88\sec^4(x)\tan^2(x) + 16\sec^6(x)
\]

Thus, the derivative of \( F(x) \) is:
\[
f'(x) = 16\sec^2(x)\tan^4(x) + 88\sec^4(x)\tan^2(x) + 16\sec^6(x)
\]

Would you like more details on any of the steps or the methods used?

### Related questions:
1. How do you apply the chain rule in differentiation?
2. What is the derivative of \( \sec(x) \) and \( \tan(x) \)?
3. Can you explain the product rule with more examples?
4. How does the chain rule apply to trigonometric functions?
5. What are some applications of derivatives involving trigonometric functions?

### Tip:
When differentiating products of trigonometric functions, always remember to use both the product and chain rules where necessary to avoid mistakes.

Learn how to differentiate the function 8sec²xtan³x + 16sec⁴xtanx step-by-step. This solution involves the product rule, chain rule, and trigonometric derivatives. Suitable for advanced high school students in Grades 11-12.

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