Question

attempt 1: 10 attempts remaining. differentiate: $f(x) = (19x^3 - 15x^6)^{5/2}$ $f(x) = $ submit answer next item

Explanation:

Step1: Apply the chain rule

The function is \( f(x) = (19x^3 - 15x^6)^{\frac{5}{2}} \). Let \( u = 19x^3 - 15x^6 \), so \( f(u)=u^{\frac{5}{2}} \). By the chain rule, \( f'(x)=f'(u)\cdot u' \). First, find \( f'(u) \): using the power rule, \( \frac{d}{du}(u^n)=nu^{n - 1} \), so \( f'(u)=\frac{5}{2}u^{\frac{5}{2}-1}=\frac{5}{2}u^{\frac{3}{2}} \).

Step2: Find \( u' \)

Differentiate \( u = 19x^3 - 15x^6 \) with respect to \( x \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we get \( u'=\frac{d}{dx}(19x^3)-\frac{d}{dx}(15x^6)=19\times3x^{2}-15\times6x^{5}=57x^{2}-90x^{5} \).

Step3: Substitute back \( u \) and \( u' \)

Substitute \( u = 19x^3 - 15x^6 \) and \( u' = 57x^{2}-90x^{5} \) into \( f'(x)=f'(u)\cdot u' \). So \( f'(x)=\frac{5}{2}(19x^3 - 15x^6)^{\frac{3}{2}}\cdot(57x^{2}-90x^{5}) \). We can factor out 3 from \( 57x^{2}-90x^{5} \) to get \( 3(19x^{2}-30x^{5}) \), then multiply by \( \frac{5}{2} \): \( \frac{5}{2}\times3(19x^3 - 15x^6)^{\frac{3}{2}}(19x^{2}-30x^{5})=\frac{15}{2}(19x^{2}-30x^{5})(19x^3 - 15x^6)^{\frac{3}{2}} \). Or we can also factor \( x^2 \) from \( 19x^{2}-30x^{5} \) and \( x^3 \) from \( 19x^3 - 15x^6 \) for further simplification, but the above form is also correct.

Answer:

\( \frac{5}{2}(57x^{2}-90x^{5})(19x^{3}-15x^{6})^{\frac{3}{2}} \) (or simplified forms like \( \frac{15}{2}(19x^{2}-30x^{5})(19x^{3}-15x^{6})^{\frac{3}{2}} \))