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Use logarithmic differentation to find the derivative of the function.

 

\(y=\sqrt{x}{e}^{{x}^{2}}({{x}^{2}+1})^{10}\)

 

Anyone who knows how to solve for the answer and can write down the steps, I would really appreciate it.  Thanks.

gibsonj338  May 22, 2018
 #1
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It uses implicit differentiation:

d/dx(y) = d/dx({e^(x^2) sqrt(x) (1 + x^2)^10})

 

The derivative of y is y'(x):

y'(x) = d/dx({e^(x^2) sqrt(x) (1 + x^2)^10})

 

Using the chain rule, d/dx({e^(x^2) sqrt(x) (x^2 + 1)^10}) = ( du)/( du) ( du)/( dx), where u = e^(x^2) sqrt(x) (x^2 + 1)^10 and d/( du)({u}) = {1}:

y'(x) = {d/dx(e^(x^2) sqrt(x) (1 + x^2)^10)}

 

Use the product rule, d/dx(u v) = v ( du)/( dx) + u ( dv)/( dx), where u = e^(x^2) and v = sqrt(x) (x^2 + 1)^10:

y'(x) = {sqrt(x) (x^2 + 1)^10 d/dx(e^(x^2)) + e^(x^2) d/dx(sqrt(x) (1 + x^2)^10)}

 

Using the chain rule, d/dx(e^(x^2)) = ( d e^u)/( du) ( du)/( dx), where u = x^2 and d/( du)(e^u) = e^u:

y'(x) = {e^(x^2) (d/dx(sqrt(x) (1 + x^2)^10)) + e^(x^2) d/dx(x^2) sqrt(x) (1 + x^2)^10}

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2.

d/dx(x^2) = 2 x:

y'(x) = {e^(x^2) (d/dx(sqrt(x) (1 + x^2)^10)) + 2 x e^(x^2) sqrt(x) (1 + x^2)^10}

 

Simplify the expression:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) (d/dx(sqrt(x) (1 + x^2)^10))}

 

Use the product rule, d/dx(u v) = v ( du)/( dx) + u ( dv)/( dx), where u = sqrt(x) and v = (x^2 + 1)^10:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + (x^2 + 1)^10 d/dx(sqrt(x)) + sqrt(x) d/dx((1 + x^2)^10) e^(x^2)}

 

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1/2.

d/dx(sqrt(x)) = d/dx(x^(1/2)) = x^(-1/2)/2:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) (sqrt(x) (d/dx((1 + x^2)^10)) + 1/(2 sqrt(x)) (1 + x^2)^10)}

 

Using the chain rule, d/dx((x^2 + 1)^10) = ( du^10)/( du) ( du)/( dx), where u = x^2 + 1 and d/( du)(u^10) = 10 u^9:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) ((1 + x^2)^10/(2 sqrt(x)) + 10 (x^2 + 1)^9 d/dx(1 + x^2) sqrt(x))}

Differentiate the sum term by term:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) ((1 + x^2)^10/(2 sqrt(x)) + d/dx(1) + d/dx(x^2) 10 sqrt(x) (1 + x^2)^9)}

 

The derivative of 1 is zero:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) ((1 + x^2)^10/(2 sqrt(x)) + 10 sqrt(x) (1 + x^2)^9 (d/dx(x^2) + 0))}

 

Simplify the expression:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) ((1 + x^2)^10/(2 sqrt(x)) + 10 sqrt(x) (1 + x^2)^9 (d/dx(x^2)))}

 

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2.

d/dx(x^2) = 2 x:

y'(x) = {2 e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) ((1 + x^2)^10/(2 sqrt(x)) + 2 x 10 sqrt(x) (1 + x^2)^9)}

 

Simplify the expression:

 

| y'(x) = {2e^(x^2) x^(3/2) (1 + x^2)^10 + e^(x^2) (20 x^(3/2) (1 + x^2)^9 + (1 + x^2)^10/(2 sqrt(x)))}  [Courtesy of Mathematica 11 Home Edition]

Guest May 22, 2018
 #2
avatar+1868 
0

This may be the right answer; however, the instructions say to use logarithmic differentation.  If anyone knows how to solve for the answer usin logarithmic differentation, I would really appreciate it.  Thanks.

gibsonj338  May 22, 2018
 #3
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It gives this answer using "Logarithmic Differentiation", but does not give step by step answer. Sorry about that:

 

Log(y'(x) = 2e^(x^2) x^(3/2) (x^2 + 1)^10 + (e^(x^2) (x^2 + 1)^10)/(2 sqrt(x)) + 20 e^(x^2) x^(3/2) (x^2 + 1)^9). Note: it uses natural logs.

Guest May 22, 2018
 #4
avatar+20598 
+1

Use logarithmic differentation to find the derivative of the function.

\(y=\sqrt{x}{e}^{({x}^{2})}({{x}^{2}+1})^{10}\)

y=\sqrt{x}{e}^{(x^2)}(x^2+1)^{10}

Anyone who knows how to solve for the answer and can write down the steps?

 

Formula

\(\begin{array}{|rclcl|} \hline \left(~\ln f(x)~ \right)' = \dfrac{f'(x)}{f(x)} \\ \hline \end{array} \)

 

1. logarithm of both sides

\(\begin{array}{|rcll|} \hline y &=& \sqrt{x}{e}^{(x^2)}(x^2+1)^{10} \quad & | \quad \text{$\ln()$ both sides} \\ \ln(y) &=& \ln(\sqrt{x})+\ln({e}^{(x^2)}) + \ln( (x^2+1)^{10} ) \\ \ln(y) &=& \ln(x^{\frac12})+\ln({e}^{(x^2)}) + \ln( (x^2+1)^{10} ) \quad & | \quad \text{Formula: $\ln(a^b) = b\ln(a) $} \\ \ln(y) &=& \frac12\ln(x)+x^2\ln(e) + 10\ln(x^2+1) \quad & | \quad \text{Formula: $\ln(e) = 1 $} \\ \ln(y) &=& \frac12\ln(x)+x^2 + 10\ln(x^2+1) \\ \hline \end{array}\)

 

2. derivation of both sides

\(\begin{array}{|rcll|} \hline \ln(y) &=& \frac12\ln(x)+x^2 + 10\ln(x^2+1) \quad & | \quad \text{derivate both sides} \\ \Big(\ln(y)\Big)' &=& \left(\frac12\ln(x) \right)' + \left(x^2 \right)' + \left(10\ln(x^2+1) \right)' \\\\ \dfrac{y'}{y} &=& \frac12\cdot \frac1x + 2x + 10\cdot \frac{2x}{x^2+1} \\\\ \dfrac{y'}{y} &=& \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \quad & | \quad \cdot y \\\\ y' &=& y\cdot \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \quad & | \quad y = \sqrt{x}e^{(x^2)}(x^2+1)^{10} \\\\ y' &=& \sqrt{x}e^{(x^2)}(x^2+1)^{10} \cdot \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \\\\ y' &=& x^{\frac12}e^{(x^2)}(x^2+1)^{10} \cdot \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \\\\ y' &=& \frac{ x^{\frac12}{e}^{(x^2)}(x^2+1)^{10} }{2x} + x^{\frac12}e^{(x^2)}(x^2+1)^{10} \cdot 2x \\ &+& x^{\frac12}e^{(x^2)}(x^2+1)^{10} \cdot \frac{20x}{x^2+1} \\\\ y' &=& \frac{ e^{(x^2)}(x^2+1)^{10} }{2x^{1-\frac12}} + 2{e}^{(x^2)}x^{\frac12+1}(x^2+1)^{10}\\ &+& 20e^{(x^2)} x^{\frac12+1}(x^2+1)^{10} \cdot \frac{1}{(x^2+1)^1} \\\\ y' &=& \frac{ e^{(x^2)}(x^2+1)^{10} }{2x^{\frac12}} + 2{e}^{(x^2)}x^{\frac32}(x^2+1)^{10}\\ &+& 20e^{(x^2)} x^{\frac32}(x^2+1)^{10-1} \\\\ y' &=& \frac{ e^{(x^2)}(x^2+1)^{10} }{2\sqrt{x}} + 2{e}^{(x^2)}x^{\frac32}(x^2+1)^{10}\\ &+& 20e^{(x^2)} x^{\frac32}(x^2+1)^{9} \\ \hline \end{array} \)

 

\(\begin{array}{rcll} \mathbf{ y' } & \mathbf{=} & \mathbf{ \dfrac{ e^{(x^2)}(x^2+1)^{10} }{2\sqrt{x}} + 2{e}^{(x^2)}x^{\frac32}(x^2+1)^{10} + 20e^{(x^2)} x^{\frac32}(x^2+1)^{9} } \\ \end{array} \)

 

laugh

heureka  May 23, 2018

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