differential calculus

[JAY6150]

Given that F(0) = 2 and F'(0) = -1, find G'(0), where G(x) = x/[1+sec(F(2x))]

 

[Deleted member 4993]

Given that F(0) = 2 and F'(0) = -1, find G'(0), where G(x) = x/1+sec(F(2x))

Using standard Law of Quotients for differentiation, calculate G'(x) and then G'(0).

Please show us what you have tried and exactly where you are stuck.

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[nasi112]

first find [MATH]G'(x)[/MATH]

 

[pka]

Given that F(0) = 2 and F'(0) = -1, find G'(0), where G(x) = x/1+sec(F(2x))

According to the image that you posted \(\large\bf G(x)=\dfrac{x}{1+\sec(F(2x))}\).

So that \(\large G'(x)=\dfrac{[1+\sec(F(2x))]-x[\tan(F(2x))\sec(F(2x))]F'(2x)(2)}{[1+\sec(F(2x))]^2}\) SEE HERE

 

[lex]

[MATH]G'(0)=\lim \limits_{h \to 0} \hspace1ex \frac{1}{1+\sec(F(2h))}=\frac{1}{1+\sec 2}[/MATH]
since F is continuous at 0 and [MATH]\sec x[/MATH] at 2.