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if f(x) +x^5[f(x)]^3=4005 and f(2) = 5, find f'(2)=
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\[f'(x)+5x^4 \cdot [f(x)]^3+x^5 \cdot 3[f(x)]^2 f'(x)=0\] now plug in x=2
and solve for f'(2)
\[f'(2)+5(2)^4 \cdot [f(2)]^3+2^5 \cdot 3 [f(2)]^2 f'(2)=0\] \[f'(2)+f'(2) \cdot 2^5 \cdot 3 [f(2)]^2=-5(2)^4 \cdot[f(2)]^3\] \[f'(2)[1+2^5 \cdot 3 [f(2)]^2=-5(2)^4 \cdot [f(2)]^3\]
\[f'(2)[1+2^5 \cdot 3 [f(2)]^2]=-5(2)^4 \cdot [f(2)]^3*\]
\[f'(2)[1+96[5]^2]=-80[5]^3\]
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\[f'(2)[1+96(25)]=-80[125]\] you got it from here?
what am i suppose to do with that?
solve for f'(2)
divde both sides by [1+96(25)]
i got -4.16 but it was wrong
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\[\frac{-10000}{2401}\]
lol thankss
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