Jawaban:
Penjelasan dengan langkah-langkah:
f(x)=6√4x-3
u(x)=(4x-3) maka u´(x)=4
f(x)=(u(x))¹/₂
f´(x)=6.(4x-3)¹/₂ .u´(x)
f´(x)=6.1(u(x))¹/₂⁻¹.u´(x)
2
=6(1(4x-3)^¹/₂⁻¹. (4)
=6(1(4x-3)⁻¹/₂.(4)
=6(4)
2(4x-3)¹/₂
= 24
2√(4x-3)
= 12 [✓]
√(4x-3)
Detail Jawaban:
Mapel : Matematika
Kelas : 11 / [XI] SMA
Materi : Fungsi dan Turunannya
Kode Kategorisasi : 11.2.7
Kata Kunci : Fungsi dan Turunannya
Demikian
Semoga bermanfaat dan bermanfaat!
Dicari turunan dari
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Jawaban:
[Fungsi dan Turunannya]
Penjelasan dengan langkah-langkah:
f(x)=6√4x-3
u(x)=(4x-3) maka u´(x)=4
f(x)=(u(x))¹/₂
f´(x)=6.(4x-3)¹/₂ .u´(x)
f´(x)=6.1(u(x))¹/₂⁻¹.u´(x)
2
=6(1(4x-3)^¹/₂⁻¹. (4)
2
=6(1(4x-3)⁻¹/₂.(4)
2
=6(4)
2(4x-3)¹/₂
= 24
2√(4x-3)
= 12 [✓]
√(4x-3)
Detail Jawaban:
Mapel : Matematika
Kelas : 11 / [XI] SMA
Materi : Fungsi dan Turunannya
Kode Kategorisasi : 11.2.7
Kata Kunci : Fungsi dan Turunannya
Demikian
Semoga bermanfaat dan bermanfaat!
Dicari turunan dari
![f(x)=6\sqrt{4x-3}\\\\(uv)' = u'v+uv'\\\\u=6,v=\sqrt{4x-3}\\\\f'(x)=\\6'\sqrt{4x-3}+6(\sqrt{4x-3})'=\\\\0+6(\sqrt{4x-3})'=\\\\6(\frac{1}{2}(4x-3)^{(1/2)-1}(4x-3)')=\\\\^{3}\not6(\frac{4}{\not2}(4x-3)^{(-1/2)})=\\\\\displaystyle\bf \frac{12}{\sqrt{4x-3}} f(x)=6\sqrt{4x-3}\\\\(uv)' = u'v+uv'\\\\u=6,v=\sqrt{4x-3}\\\\f'(x)=\\6'\sqrt{4x-3}+6(\sqrt{4x-3})'=\\\\0+6(\sqrt{4x-3})'=\\\\6(\frac{1}{2}(4x-3)^{(1/2)-1}(4x-3)')=\\\\^{3}\not6(\frac{4}{\not2}(4x-3)^{(-1/2)})=\\\\\displaystyle\bf \frac{12}{\sqrt{4x-3}}](https://tex.z-dn.net/?f=f%28x%29%3D6%5Csqrt%7B4x-3%7D%5C%5C%5C%5C%28uv%29%27%20%3D%20u%27v%2Buv%27%5C%5C%5C%5Cu%3D6%2Cv%3D%5Csqrt%7B4x-3%7D%5C%5C%5C%5Cf%27%28x%29%3D%5C%5C6%27%5Csqrt%7B4x-3%7D%2B6%28%5Csqrt%7B4x-3%7D%29%27%3D%5C%5C%5C%5C0%2B6%28%5Csqrt%7B4x-3%7D%29%27%3D%5C%5C%5C%5C6%28%5Cfrac%7B1%7D%7B2%7D%284x-3%29%5E%7B%281%2F2%29-1%7D%284x-3%29%27%29%3D%5C%5C%5C%5C%5E%7B3%7D%5Cnot6%28%5Cfrac%7B4%7D%7B%5Cnot2%7D%284x-3%29%5E%7B%28-1%2F2%29%7D%29%3D%5C%5C%5C%5C%5Cdisplaystyle%5Cbf%20%5Cfrac%7B12%7D%7B%5Csqrt%7B4x-3%7D%7D)