Question

attempt 1: 10 attempts remaining. given ( f(x) = (5x^3 - 9x^2 + 6x - 4)^{10} ), find ( f(x) ). ( f(x) = ) input box submit answer next item

Explanation:

Step1: Apply the chain rule

The chain rule states that if \( y = u^n \) where \( u \) is a function of \( x \), then \( y' = n u^{n - 1} \cdot u' \). Let \( u = 5x^3 - 9x^2 + 6x - 4 \) and \( n = 10 \). First, find the derivative of \( u \) with respect to \( x \).
The derivative of \( u = 5x^3 - 9x^2 + 6x - 4 \) is \( u'=\frac{d}{dx}(5x^3)-\frac{d}{dx}(9x^2)+\frac{d}{dx}(6x)-\frac{d}{dx}(4) \).
Using the power rule \( \frac{d}{dx}(ax^n)=nax^{n - 1} \) and the derivative of a constant is 0:
\( \frac{d}{dx}(5x^3)=15x^2 \), \( \frac{d}{dx}(9x^2) = 18x \), \( \frac{d}{dx}(6x)=6 \), \( \frac{d}{dx}(4) = 0 \). So \( u'=15x^2-18x + 6 \).

Step2: Apply the chain rule formula

Now, using the chain rule \( f'(x)=n u^{n - 1}\cdot u' \), where \( n = 10 \), \( u = 5x^3 - 9x^2 + 6x - 4 \) and \( u'=15x^2-18x + 6 \).
So \( f'(x)=10(5x^3 - 9x^2 + 6x - 4)^{9}\cdot(15x^2-18x + 6) \). We can factor out a 3 from the second factor: \( 15x^2-18x + 6=3(5x^2 - 6x + 2) \). Then \( f'(x)=10\times3(5x^3 - 9x^2 + 6x - 4)^{9}(5x^2 - 6x + 2)=30(5x^3 - 9x^2 + 6x - 4)^{9}(5x^2 - 6x + 2) \) (or we can leave it as \( 10(5x^3 - 9x^2 + 6x - 4)^{9}(15x^2-18x + 6) \)).

Answer:

\( 10(5x^3 - 9x^2 + 6x - 4)^9(15x^2 - 18x + 6) \) (or simplified as \( 30(5x^3 - 9x^2 + 6x - 4)^9(5x^2 - 6x + 2) \))