Question

1. 0 / 0.62 points

practice another
differentiate.
\\( y = 4x^2 \sin x \tan x \\)
\\( y = \\)
check answer

Explanation:

Step1: Apply Product Rule

The function \( y = 4x^2\sin x\tan x \) is a product of three functions: \( u = 4x^2 \), \( v=\sin x \), \( w = \tan x \). The product rule for three functions is \( (uvw)'=u'vw + uv'w+ uvw' \).
First, find \( u' \): \( u = 4x^2 \), so \( u'=8x \).
Find \( v' \): \( v=\sin x \), so \( v'=\cos x \).
Find \( w' \): \( w = \tan x \), so \( w'=\sec^2 x \).

Step2: Substitute into Product Rule

Substitute \( u, u', v, v', w, w' \) into the formula:
\( y'=u'vw + uv'w+ uvw' \)
\( = 8x\cdot\sin x\cdot\tan x+4x^2\cdot\cos x\cdot\tan x + 4x^2\cdot\sin x\cdot\sec^2 x \)

Simplify each term:
- For the second term: \( \tan x=\frac{\sin x}{\cos x} \), so \( 4x^2\cdot\cos x\cdot\frac{\sin x}{\cos x}=4x^2\sin x \)
- For the third term: \( \sec^2 x=\frac{1}{\cos^2 x} \), so \( 4x^2\cdot\sin x\cdot\frac{1}{\cos^2 x}=4x^2\frac{\sin x}{\cos^2 x} = 4x^2\sin x\sec^2 x \) (or we can also write it as \( 4x^2\tan x\sec x \) since \( \frac{\sin x}{\cos^2 x}=\tan x\sec x \))
- The first term remains \( 8x\sin x\tan x \)

So combining and simplifying:
\( y' = 8x\sin x\tan x+4x^2\sin x + 4x^2\sin x\sec^2 x \)
We can also factor out common terms. Notice that \( \tan x=\frac{\sin x}{\cos x} \) and \( \sec^2 x = 1 + \tan^2 x \), but another way:
From the original substitution, after simplifying the second term, we can also rewrite the expression as:
\( y'=8x\sin x\tan x + 4x^2\sin x(1 + \sec^2 x) \) (since the second term is \( 4x^2\sin x \) and the third term is \( 4x^2\sin x\sec^2 x \), factor out \( 4x^2\sin x \) from the last two terms)

Or, let's do the simplification more directly:

First term: \( 8x\sin x\tan x \)
Second term: \( 4x^2\cos x\tan x=4x^2\cos x\cdot\frac{\sin x}{\cos x}=4x^2\sin x \)
Third term: \( 4x^2\sin x\sec^2 x \)

So the final simplified form (we can also combine terms with \( \sin x \)):
\( y' = 8x\sin x\tan x+4x^2\sin x(1 + \sec^2 x) \)
Or, if we want to express everything in terms of \( \sin x \) and \( \cos x \):
\( \tan x=\frac{\sin x}{\cos x} \), \( \sec^2 x=\frac{1}{\cos^2 x} \)
So:
\( y' = 8x\sin x\cdot\frac{\sin x}{\cos x}+4x^2\sin x + 4x^2\sin x\cdot\frac{1}{\cos^2 x} \)
\( = \frac{8x\sin^2 x}{\cos x}+4x^2\sin x+\frac{4x^2\sin x}{\cos^2 x} \)
We can factor out \( \frac{4x\sin x}{\cos^2 x} \):
\( = \frac{4x\sin x}{\cos^2 x}(2x\sin x\cos x + x\cos x + x) \) Wait, maybe that's more complicated. The standard form after applying product rule and simplifying the second term is:

\( y' = 8x\sin x\tan x + 4x^2\sin x + 4x^2\sin x\sec^2 x \)

Alternatively, we can also write it as:

First, recall that \( \tan x = \frac{\sin x}{\cos x} \), so let's rewrite the original function: \( y = 4x^2\frac{\sin^2 x}{\cos x} \) (since \( \sin x\tan x=\sin x\cdot\frac{\sin x}{\cos x}=\frac{\sin^2 x}{\cos x} \))
Now, use quotient rule: if \( y=\frac{f(x)}{g(x)} \), then \( y'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} \), where \( f(x)=4x^2\sin^2 x \), \( g(x)=\cos x \)
Find \( f'(x) \): use product rule on \( 4x^2\sin^2 x \), let \( a = 4x^2 \), \( b=\sin^2 x \), so \( f'(x)=a'b + ab' \), \( a' = 8x \), \( b' = 2\sin x\cos x \), so \( f'(x)=8x\sin^2 x+4x^2\cdot 2\sin x\cos x=8x\sin^2 x + 8x^2\sin x\cos x \)
\( g'(x)=-\sin x \)
Then \( y'=\frac{(8x\sin^2 x + 8x^2\sin x\cos x)\cos x-4x^2\sin^2 x(-\sin x)}{\cos^2 x} \)
\( =\frac{8x\sin^2 x\cos x + 8x^2\sin x\cos^2 x + 4x^2\sin^3 x}{\cos^2 x} \)
Factor out \( 4x\sin x \) from numerator:
\( =\frac{4x\sin x(2\sin x\cos x + 2x\cos^2 x + x\sin^2 x)}{\cos^2 x} \)
But this seems more complicated. Going back to the product rule result:
\( y' = 8x\sin x\tan x+4x^2\sin x + 4x^2\sin x\sec^2 x \)
We can factor \( 4x\sin x \) from the first and third term? Wait, first term: \( 8x\sin x\tan x = 4x\sin x\cdot 2\tan x \), third term: \( 4x^2\sin x\sec^2 x=4x\sin x\cdot x\sec^2 x \), second term: \( 4x^2\sin x = 4x\sin x\cdot x \)
Alternatively, the simplest form from product rule is:
\( y' = 8x\sin x\tan x + 4x^2\sin x + 4x^2\sin x\sec^2 x \)
Or, combining like terms for the terms with \( \sin x \):
\( y' = 8x\sin x\tan x + 4x^2\sin x(1 + \sec^2 x) \)
Since \( 1+\sec^2 x = 1 + 1+\tan^2 x=2 + \tan^2 x \), but maybe not necessary. The key is that after applying the product rule correctly, we get the derivative.

Answer:

\( y' = 8x\sin x\tan x + 4x^2\sin x + 4x^2\sin x\sec^2 x \) (or equivalent simplified forms, e.g., \( y' = 8x\sin x\tan x + 4x^2\sin x(1 + \sec^2 x) \) or further simplified using trigonometric identities)