What is the derivative of (x^2)(sinx)(tanx) without using the chain rule?

Question

Christopher Allers · Accepted Answer

You can manipulate your function remembering that:
#tan(x)=sin(x)/cos(x)#
and #sin^2(x)=1-cos^2(x)#
and get:

Ezra Adams · Answer

Use the product rule for three factors:

#d/(dx)(fgh) = f'gh+fg'h+fgh'#

For #y = x^2sinxtanx#, we get

#y' = 2xsinxtanx + x^2 cosxtanx+x^2sinxsec^2x#

We ca rewrite the middle term more simply, and we may choose to rewrite the third term:

#y' = 2xsinxtanx + x^2 sinx+x^2tanxsecx#.

#d/(dx)(fgh) =f'[gh]+f d/(dx)[gh]#

#color(white)"sssssssss"# #=f'gh+f[g'h+gh']#

#color(white)"sssssssss"# #=f'gh+fg'h+fgh'#.

Paisley Johnson · Answer

undefined To find the derivative of $x^2 \cdot \sin(x) \cdot \tan(x)$ without using the chain rule, you can use the product rule and trigonometric identities. First, expand the expression:

$$y = x^2 \cdot \sin(x) \cdot \tan(x)$$

Now, apply the product rule:

$$y' = (x^2)' \cdot \sin(x) \cdot \tan(x) + x^2 \cdot (\sin(x))' \cdot \tan(x) + x^2 \cdot \sin(x) \cdot (\tan(x))'$$

Differentiate each term using basic differentiation rules and trigonometric identities:

$$(x^2)' = 2x$$
$$(\sin(x))' = \cos(x)$$
$$(\tan(x))' = \sec^2(x)$$

Substitute the derivatives back into the expression:

$$y' = 2x \cdot \sin(x) \cdot \tan(x) + x^2 \cdot \cos(x) \cdot \tan(x) + x^2 \cdot \sin(x) \cdot \sec^2(x)$$

This is the derivative of $x^2 \cdot \sin(x) \cdot \tan(x)$ without using the chain rule.