f'(x) = ____
Gunakan " Definisi Limit "
[tex]f'(x) = \bf21x^2-6x+20x^3[/tex]
Pembahasan
Turunan Berdasarkan Definisi
Diberikan: [tex]f(x)=7x^3-3x^2+5x^4[/tex]
Turunannya:
[tex]\begin{aligned}&f'(x)\\&{=\ }\lim_{h\to0}\:\frac{f(x+h)-f(x)}{h}\\&{=\ }\lim_{h\to0}\:\frac{1}{h}\Bigl[f(x+h)-f(x)\Bigr]\\&{=\ }\lim_{h\to0}\:\frac{1}{h}\left[\begin{array}{c}7(x+h)^3-3(x+h)^2+5(x+h)^4\\-\ \left(7x^3-3x^2+5x^4\right)\end{array}\right]\\&{=\ }\lim_{h\to0}\:\frac{1}{h}\left[\begin{array}{c}(x+h)^2\left(7(x+h)-3+5(x+h)^2\right)\\-\ x^2\left(7x-3+5x^2\right)\end{array}\right]\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\lim_{h\to0}\:\frac{1}{h}\left[\begin{array}{c}\left(x^2+2hx+h^2\right)\left[7x+h-3+5\left(x^2+2hx+h^2\right)\right]\\-\ x^2\left(7x-3+5x^2\right)\end{array}\right]\\&{=\ }\lim_{h\to0}\:\frac{1}{h}\left[\begin{array}{c}\left(x^2+2hx+h^2\right)\left[7x+7h-3+5x^2+10hx+5h^2\right]\\-\ x^2\left(7x-3+5x^2\right)\end{array}\right]\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\lim_{h\to0}\:\frac{1}{h}\left[\begin{array}{c}x^2\left(\cancel{7x}+7h-\cancel{3}+\cancel{5x^2}+10hx+5h^2-\cancel{7x}+\cancel{3}-\cancel{5x^2}\right)\\+\ 2hx\left(7x+7h-3+5x^2+10hx+5h^2\right)\\+\ h^2\left(7x+7h-3+5x^2+10hx+5h^2\right)\end{array}\right]\\&{=\ }\lim_{h\to0}\:\frac{1}{h}\left[\begin{array}{c}x^2\left(7h+10hx+5h^2\right)\\+\ 2hx\left(7x+7h-3+5x^2+10hx+5h^2\right)\\+\ h^2\left(7x+7h-3+5x^2+10hx+5h^2\right)\end{array}\right]\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\lim_{h\to0}\:\left[\begin{array}{c}x^2\left(7+10x+5h\right)\\+\ 2x\left(7x+7h-3+5x^2+10hx+5h^2\right)\\+\ h\left(7x+7h-3+5x^2+10hx+5h^2\right)\end{array}\right]\\&{=\ }x^2\left(7+10x+5(0)\right)\\&{\quad}+2x\left(7x+7(0)-3+5x^2+10(0)x+5(0^2)\right)\\&{\quad}+0\left(7x+7h-3+5x^2+10hx+5h^2\right)\\&{=\ }7x^2+10x^3+14x^2-6x+10x^3\\&{=\ }20x^3+21x^2-6x\\&{=\ }\boxed{\ \bf21x^2-6x+20x^3\ }\end{aligned}[/tex]
Penyelesaian:
Definisi limit
Lim h→0 f(x+h)-f(x)/h
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f(x)=5x⁴+7x³-3x²
f(x+h)=5(x+h)⁴+7(x+h)³-3(x+h)²
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(x+h)(x+h)
=x²+2xh+h²
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(x+h)(x+h)(x+h)
=(x²+2xh+h²)(x+h)
=x³+x²h+2x²h+2xh²+xh²+h³
=x³+3x²h+3xh²+h³
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(x+h)(x+h)(x+h)(x+h)
=(x²+2xh+h²)(x+h)(x+h)
=(x³+x²h+2x²h+2xh²+xh²+h³)(x+h)
=(x³+3x²h+3xh²+h³)(x+h)
=(x⁴+x³h+3x³h+3x²h²+3x²h²+3xh³+xh³+h⁴)
=x⁴+4x³h+6x²h²+4xh³+h⁴
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f(x+h)=5(x⁴+4x³h+6x²h²+4xh³+h⁴)+7(x³+3x²h+3xh²+h³)-3(x²+2xh+h²)
f(x+h)=5x⁴+20x³h+30x²h²+20xh³+5h⁴+7x³+21x²h+21xh²+7h³-3x²-6xh-3h²
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Substitusikan
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Lim h→0 f(x+h)-f(x)/h
=Lim h→0 5x⁴+20x³h+30x²h²+20xh³+5h⁴+7x³+21x²h+21xh²+7h³-3x²-6xh-3h²-(5x⁴+7x³-3x²)/h
=Lim h→0 20x³h+30x²h²+20xh³+5h⁴+21x²h+21xh²+7h³-6xh-3h²/h
=Lim h→0 h(20x³+30x²h+20xh²+5h³+21x²+21xh+7h²-6x-3h)/h
=Lim h→0 20x³+30x²h+20xh²+5h³+21x²+21xh+7h²-6x-3h
=20x³+21x²-6x
[tex] \boxed{ \colorbox{black}{ \sf{ \color{lightgreen}{ answered\:by\:Duone}}}} [/tex]
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