Question

Brainly ,Aryabhata!!!!

Find the derivative ,BY FIRST PRINCIPAL!

i) 5secx + 4cosx
Step by step explanation
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Answer 1

Explanation:

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Let, f(x) = 5secx + 4cosx,

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According to the 1st Principal,

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$\bull \boxed{ \tt \red{f'(x) = lim_{ h \rightarrow0}( \dfrac{f(x + h) - f(x)}{h} ) } }$

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$\leadsto \: \tt \: f'(x) = lim_{h \rightarrow0} \dfrac{5sec(x + h) + 4cos(x + h) - (5secx +4 cosx)}{h}$

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$\leadsto \: \tt \: f'(x) = 5 lim_{h \rightarrow0} \dfrac{sec(x + h) - (secx)}{h} + 4lim_{h \rightarrow0} \dfrac{cos(x + h) - (cosx)}{h}$

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$\leadsto \: \tt \: f'(x) = 5 lim_{h \rightarrow0} \dfrac{1}{h} ( \dfrac{1}{cos(x + h)} - \frac{1}{cosx} ) + 4lim_{h \rightarrow0} \dfrac{1}{h} (cos(x + h) - cosx)$

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$\leadsto \: \tt \: f'(x) = 5 lim_{h \rightarrow0} ( \dfrac{cosx - cos(x + h)}{cosx.cos(x + h)} ) + 4 lim_{h \rightarrow0} \dfrac{1}{h} (cosx cosh - sinx sinh - cosx)$

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$\leadsto \: \tt \: f'(x) = \dfrac{5}{cosx} lim_{h \rightarrow0} \dfrac{1}{h} ( \dfrac{ - 2sin( \dfrac{x + x + h}{2})sin( \dfrac{ \cancel{x - x} - h}{2} ) }{ cos(x + h) } ) + 4 lim_{h \rightarrow0} \dfrac{1}{h} ( - cosx( 1- cosh) - sinx.sinh)$

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$\leadsto \: \tt \: f'(x) = \dfrac{5}{cosx} lim_{h \rightarrow0} \dfrac{1}{h} ( \dfrac{ - 2sin( \dfrac{2x+ h}{2})sin( \dfrac{ - h}{2} ) }{ cos(x + h) } ) + 4( - cosx \: lim_{ h \rightarrow0}( \dfrac{1 - cosh}{h} ) - sinx \: lim_{h \rightarrow0} \dfrac{sinh}{h} )$

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$\leadsto \: \tt \: f'(x) = \dfrac{5}{cosx} lim_{h \rightarrow0} \dfrac{1}{h}( \dfrac{ sin( \dfrac{2x + h}{2} ). \dfrac{sin( \dfrac{h}{2}) }{ \dfrac{h}{2} } }{cos(x + h)}) + 4( - cosx.0 - sinx.1)$

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$\leadsto \: \tt \: f'(x) = \dfrac{5}{cosx}( lim_{h \rightarrow0} \dfrac{sinx \dfrac{(2x + h)}{2}}{cos(x + h)} . lim_{h \rightarrow0}( \dfrac{sin( \dfrac{h}{2} )}{( \dfrac{h}{2} )} ) - 4sinx$

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$\leadsto \: \tt \: f'(x) = \dfrac{5}{cosx} . \dfrac{sinx}{cosx} .1 - 4sinx$

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$\leadsto \: \tt \: f'(x) = 5 \dfrac{1}{cosx} .tanx - 4sinx$

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$\leadsto \: \underline{\boxed{\tt \green{ f'(x) = 5secx .tanx - 4sinx}}}$