Question

attempt 1: 10 attempts remaining. differentiate: ( f(x) = (17x^3 - 15x^6)^{7/2} ) ( f(x) = ) input box submit answer next item

Explanation:

Step1: Identify the outer and inner functions

The function \( f(x) = (17x^3 - 15x^6)^{\frac{7}{2}} \) is a composite function. Let \( u = 17x^3 - 15x^6 \) (inner function) and \( y = u^{\frac{7}{2}} \) (outer function). We will use the chain rule, which states that \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \).

Step2: Differentiate the outer function with respect to \( u \)

For \( y = u^{\frac{7}{2}} \), using the power rule \( \frac{d}{du}(u^n)=nu^{n - 1} \), we have:
\( \frac{dy}{du}=\frac{7}{2}u^{\frac{7}{2}-1}=\frac{7}{2}u^{\frac{5}{2}} \)

Step3: Differentiate the inner function with respect to \( x \)

For \( u = 17x^3 - 15x^6 \), using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \) for each term:
\( \frac{du}{dx}=\frac{d}{dx}(17x^3)-\frac{d}{dx}(15x^6)=17\times3x^{2}-15\times6x^{5}=51x^{2}-90x^{5} \)

Step4: Apply the chain rule

Substitute \( u = 17x^3 - 15x^6 \), \( \frac{dy}{du}=\frac{7}{2}(17x^3 - 15x^6)^{\frac{5}{2}} \) and \( \frac{du}{dx}=51x^{2}-90x^{5} \) into the chain rule formula \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \):
\( f^{\prime}(x)=\frac{7}{2}(17x^3 - 15x^6)^{\frac{5}{2}}\cdot(51x^{2}-90x^{5}) \)
We can factor out 3 from \( 51x^{2}-90x^{5} \): \( 51x^{2}-90x^{5}=3(17x^{2}-30x^{5}) \)
Then \( f^{\prime}(x)=\frac{7}{2}(17x^3 - 15x^6)^{\frac{5}{2}}\cdot3(17x^{2}-30x^{5})=\frac{21}{2}(17x^{2}-30x^{5})(17x^3 - 15x^6)^{\frac{5}{2}} \)
We can also factor \( x^2 \) from \( 17x^{2}-30x^{5} \) and \( x^3 \) from \( 17x^3 - 15x^6 \):
\( 17x^{2}-30x^{5}=x^{2}(17 - 30x^{3}) \)
\( 17x^3 - 15x^6=x^{3}(17 - 15x^{3}) \)
But the simplified form using the chain rule directly is also acceptable. The most straightforward form after applying the chain rule is:
\( f^{\prime}(x)=\frac{7}{2}(17x^{3}-15x^{6})^{\frac{5}{2}}(51x^{2}-90x^{5}) \)
Or we can simplify the coefficient and the polynomial:
\( 51x^{2}-90x^{5}=3x^{2}(17 - 30x^{3}) \)
\( \frac{7}{2}\times3x^{2}(17 - 30x^{3})(17x^{3}-15x^{6})^{\frac{5}{2}}=\frac{21}{2}x^{2}(17 - 30x^{3})(17x^{3}-15x^{6})^{\frac{5}{2}} \)
Another way is to factor \( x^3 \) from \( 17x^3 - 15x^6 \): \( 17x^3 - 15x^6=x^{3}(17 - 15x^{3}) \), then \( (17x^3 - 15x^6)^{\frac{5}{2}}=x^{\frac{15}{2}}(17 - 15x^{3})^{\frac{5}{2}} \)
And \( 51x^{2}-90x^{5}=3x^{2}(17 - 30x^{3}) \)
Multiplying together: \( \frac{7}{2}\times3x^{2}\times x^{\frac{15}{2}}(17 - 30x^{3})(17 - 15x^{3})^{\frac{5}{2}}=\frac{21}{2}x^{\frac{19}{2}}(17 - 30x^{3})(17 - 15x^{3})^{\frac{5}{2}} \)
But the simplest form using the chain rule steps is:

First, apply chain rule:

Let \( u = 17x^3-15x^6 \), \( y = u^{7/2} \)

\( y^\prime=\frac{7}{2}u^{5/2}\cdot u^\prime \)

\( u^\prime = 51x^2 - 90x^5 \)

So \( f^\prime(x)=\frac{7}{2}(17x^3 - 15x^6)^{\frac{5}{2}}(51x^2 - 90x^5) \)

We can factor 3 from \( 51x^2 - 90x^5 \):

\( 51x^2 - 90x^5=3(17x^2 - 30x^5) \)

Then \( f^\prime(x)=\frac{7}{2}\times3(17x^2 - 30x^5)(17x^3 - 15x^6)^{\frac{5}{2}}=\frac{21}{2}(17x^2 - 30x^5)(17x^3 - 15x^6)^{\frac{5}{2}} \)

Or factor \( x^2 \) from \( 17x^2 - 30x^5 \): \( 17x^2 - 30x^5=x^2(17 - 30x^3) \)

And factor \( x^3 \) from \( 17x^3 - 15x^6 \): \( 17x^3 - 15x^6=x^3(17 - 15x^3) \)

Then \( (17x^3 - 15x^6)^{\frac{5}{2}}=x^{\frac{15}{2}}(17 - 15x^3)^{\frac{5}{2}} \)

\( 17x^2 - 30x^5=x^2(17 - 30x^3) \)

Multiplying:

\( \frac{21}{2}\times x^2\times x^{\frac{15}{2}}(17 - 30x^3)(17 - 15x^3)^{\frac{5}{2}}=\frac{21}{2}x^{\frac{19}{2}}(17 - 30x^3)(17 - 15x^3)^{\frac{5}{2}} \)

But the basic form after applying the chain rule is \( f^\prime(x)=\frac{7}{2}(17x^3 - 15x^6)^{\frac{5}{2}}(51x^2 - 90x^5) \)

Answer:

\( \boldsymbol{\frac{7}{2}(17x^3 - 15x^6)^{\frac{5}{2}}(51x^2 - 90x^5)} \) (or simplified forms like \( \frac{21}{2}(17x^2 - 30x^5)(17x^3 - 15x^6)^{\frac{5}{2}} \) etc.)