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Question Number 69763 by Rio Michael last updated on 27/Sep/19

f i n d \frac{d y}{d x} i f y = 3^{x} e^{2 x + 1}, a t x = 1

Commented by mathmax by abdo last updated on 16/Oct/19

y (x) = e^{x l n 3 + 2 x + 1} \Rightarrow y^{^{'}} (x) = (l n 3 + 2) e^{(2 + l n 3) x + 1} \Rightarrow

y^{^{'}} (1) = (2 + l n 3) e^{2 + l n 3 + 1}

Commented by $@ty@m123 last updated on 16/Oct/19

I t h a s m i n o r t y p o e r r o r .

Answered by $@ty@m123 last updated on 27/Sep/19

y = 3^{x} e^{2 x + 1}

T a k e n a t u r a l l o g,

\ln y = \ln 3^{x} e^{2 x + 1}

\ln y = \ln 3^{x} + \ln e^{2 x + 1}

\ln y = x \ln 3 + 2 x + 1

D i f f e r e n t i a t i n g,

\frac{1}{y} \frac{d y}{d x} = \ln 3 + 2

\frac{d y}{d x} = y (2 + \ln 3) \dots (1)

W h e n x = 1,

y = 3^{1} . e^{3}

∴ \frac{d y}{d x} = 3 e^{3} (2 + \ln 3)

Commented by Rio Michael last updated on 27/Sep/19

t h a n k s s i r