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Question Number 122216 by help last updated on 14/Nov/20

Answered by liberty last updated on 14/Nov/20

 ℓn f(x) = arc sin 4x    ((f ′(x))/(f(x))) = (4/( (√(1+16x^2 )))) ⇒ f ′(x)= ((4e^(arc sin 4x) )/( (√(1+16x^2 ))))   f ′((1/8)) = ((4e^(π/6) )/( (√(1+(1/4))))) = ((8e^(π/6) )/( (√5))) .▲

$$\:\ell\mathrm{n}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{arc}\:\mathrm{sin}\:\mathrm{4x}\: \\ $$$$\:\frac{\mathrm{f}\:'\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{1}+\mathrm{16x}^{\mathrm{2}} }}\:\Rightarrow\:\mathrm{f}\:'\left(\mathrm{x}\right)=\:\frac{\mathrm{4e}^{\mathrm{arc}\:\mathrm{sin}\:\mathrm{4x}} }{\:\sqrt{\mathrm{1}+\mathrm{16x}^{\mathrm{2}} }} \\ $$$$\:\mathrm{f}\:'\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\:=\:\frac{\mathrm{4e}^{\frac{\pi}{\mathrm{6}}} }{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}}\:=\:\frac{\mathrm{8e}^{\frac{\pi}{\mathrm{6}}} }{\:\sqrt{\mathrm{5}}}\:.\blacktriangle\: \\ $$

Commented by help last updated on 14/Nov/20

last step please.  explain

$${last}\:{step}\:{please}. \\ $$$${explain} \\ $$

Commented by liberty last updated on 15/Nov/20

f ′((1/8)) = ((4e^(arc sin ((4/8))) )/( (√(1+((16)/(64)))))) = ((4e^(arc sin ((1/2))) )/( (√(1+(1/4)))))   = ((4e^(π/6) )/( (√(5/4)))) = ((4e^(π/6) )/((1/2)(√5))) = ((8e^(π/6) )/( (√5)))

$$\mathrm{f}\:'\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\:=\:\frac{\mathrm{4e}^{\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{4}}{\mathrm{8}}\right)} }{\:\sqrt{\mathrm{1}+\frac{\mathrm{16}}{\mathrm{64}}}}\:=\:\frac{\mathrm{4e}^{\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$$$\:=\:\frac{\mathrm{4e}^{\frac{\pi}{\mathrm{6}}} }{\:\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}}\:=\:\frac{\mathrm{4e}^{\frac{\pi}{\mathrm{6}}} }{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{8e}^{\frac{\pi}{\mathrm{6}}} }{\:\sqrt{\mathrm{5}}} \\ $$

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