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Elden [556K]
3 years ago
7

Find the linearization L(x) of y=e8xln(x) at a=1

Mathematics
1 answer:
motikmotik3 years ago
5 0
Answer: L(x) = e^8 (x-1)

Explanation:


The linearization L(x) of y = f(x) at x = a is given by

L(x) = f(a) + f'(a) (x - a)     (1)

Where f'(a) is the derivative of y = f(x) evaluated at x = a. So, we need to find the derivative of y = f(x) and evaluate the derivative at x = a. 

The derivative of y = f(x) is computed using the product rule for the derivatives because f(x) is the product of two functions: logarithmic and exponential. 

So, the derivative of y = f(x) is given by

f'(x) =  (\frac{d}{dx} (e^{8x}))(\ln x) + (\frac{d}{dx} (\ln x))(e^{8x})
\\
\\ f'(x) = 8e^{8x}\ln x + (e^{8x})(\frac{1}{x})
\\
\\\boxed {f'(x) = 8e^{8x}\ln x + \frac{e^{8x}}{x}}

Since a = 1, the derivative of y = f(x) evaluated at x = a = 1 is given by

f'(a) = f'(1) 
\\ f'(a) = 8e^{8(1)}\ln 1 + \frac{e^{8(1)}}{1}
\\
\\ f'(a) = 8e^{8}(0) + e^{8}
\\ \boxed{f'(a) = e^8}

Moreover, note that

f(a) = f(1)
\\ f(a) = e^{8(1)}(\ln 1) 
\\ f(a) = e^8 (0) 
\\ \boxed{f(a) = 0}

Using equation (1), the linearization L(x) of y = f(x) at x = a = 1 is given by

L(x) = f(a) + f'(a) (x - a)
\\ L(x) = 0 + e^8 (x - 1)
\\ \boxed{L(x) = e^8 (x - 1)}
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