Hills Crawford
12/05/2023 · Junior High School
Use the chain rule to find the derivative of \[ f(x)=8 \sqrt{6 x^{3}+8 x^{5}} \] Type your answer without fractional or negative exponents. \( f^{\prime}(x)= \)
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Step-by-step Solution
Find the first order derivative with respect to \( x \) for \( 8 \sqrt{6 x^{3}+8 x^{5}} \).
Evaluate the derivative by following steps:
- step0: Evaluate the derivative:
\(\frac{d}{dx}\left(8\sqrt{6x^{3}+8x^{5}}\right)\)
- step1: Simplify the root:
\(\frac{d}{dx}\left(8x\sqrt{6x+8x^{3}}\right)\)
- step2: Simplify:
\(\frac{d}{dx}\left(8x\left(6x+8x^{3}\right)^{\frac{1}{2}}\right)\)
- step3: Use differentiation rules:
\(\frac{d}{dx}\left(8x\right)\times \left(6x+8x^{3}\right)^{\frac{1}{2}}+8x\times \frac{d}{dx}\left(\left(6x+8x^{3}\right)^{\frac{1}{2}}\right)\)
- step4: Calculate:
\(8\left(6x+8x^{3}\right)^{\frac{1}{2}}+8x\times \frac{d}{dx}\left(\left(6x+8x^{3}\right)^{\frac{1}{2}}\right)\)
- step5: Calculate:
\(8\left(6x+8x^{3}\right)^{\frac{1}{2}}+8x\times \frac{3+12x^{2}}{\left(6x+8x^{3}\right)^{\frac{1}{2}}}\)
- step6: Calculate:
\(8\left(6x+8x^{3}\right)^{\frac{1}{2}}+\frac{24x+96x^{3}}{\left(6x+8x^{3}\right)^{\frac{1}{2}}}\)
- step7: Reduce fractions to a common denominator:
\(\frac{8\left(6x+8x^{3}\right)^{\frac{1}{2}}\sqrt{2}\times \left(3x+4x^{3}\right)^{\frac{1}{2}}}{\sqrt{2}\times \left(3x+4x^{3}\right)^{\frac{1}{2}}}+\frac{24x+96x^{3}}{\left(6x+8x^{3}\right)^{\frac{1}{2}}}\)
- step8: Calculate:
\(\frac{8\sqrt{2}\times \left(6x+8x^{3}\right)^{\frac{1}{2}}\left(3x+4x^{3}\right)^{\frac{1}{2}}}{\sqrt{2}\times \left(3x+4x^{3}\right)^{\frac{1}{2}}}+\frac{24x+96x^{3}}{\sqrt{2}\times \left(3x+4x^{3}\right)^{\frac{1}{2}}}\)
- step9: Transform the expression:
\(\frac{8\sqrt{2}\times \left(6x+8x^{3}\right)^{\frac{1}{2}}\left(3x+4x^{3}\right)^{\frac{1}{2}}+24x+96x^{3}}{\sqrt{2}\times \left(3x+4x^{3}\right)^{\frac{1}{2}}}\)
- step10: Rewrite the expression:
\(\frac{8\sqrt{2}\times \left(\left(6x+8x^{3}\right)\left(3x+4x^{3}\right)\right)^{\frac{1}{2}}+24x+96x^{3}}{\sqrt{2}\times \left(3x+4x^{3}\right)^{\frac{1}{2}}}\)
- step11: Simplify:
\(\frac{48x+2^{6}x^{3}+24x+96x^{3}}{\sqrt{2}\times \left(3x+4x^{3}\right)^{\frac{1}{2}}}\)
- step12: Simplify:
\(\frac{48x+2^{6}x^{3}+24x+96x^{3}}{\sqrt{2}\times \sqrt{3x+4x^{3}}}\)
- step13: Calculate:
\(\frac{\left(8\sqrt{x}\times \left(6+8x^{2}\right)+24\sqrt{x}+96x^{2}\sqrt{x}\right)\sqrt{x}}{\sqrt{6+8x^{2}}\times \sqrt{x}}\)
- step14: Reduce the fraction:
\(\frac{8\sqrt{x}\times \left(6+8x^{2}\right)+24\sqrt{x}+96x^{2}\sqrt{x}}{\sqrt{6+8x^{2}}}\)
- step15: Calculate:
\(\frac{36\sqrt{6x+8x^{3}}+80x^{2}\sqrt{6x+8x^{3}}}{3+4x^{2}}\)
The derivative of \( f(x)=8 \sqrt{6 x^{3}+8 x^{5}} \) using the chain rule is:
\[ f'(x) = \frac{36\sqrt{6x+8x^{3}}+80x^{2}\sqrt{6x+8x^{3}}}{3+4x^{2}} \]
Quick Answer
\[ f'(x) = \frac{36\sqrt{6x+8x^{3}}+80x^{2}\sqrt{6x+8x^{3}}}{3+4x^{2}} \]
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