Exercise
Find the derivative of the following function:
f ( x ) = ( x + 1 x ) 10 f(x)={(\sqrt{x}+\frac{1}{\sqrt{x}})}^{10} f ( x ) = ( x + x 1 ) 1 0
Final Answer
Show final answer
f'(x)=\frac{5{(x+1)}^9(x-1)}{x^6}
Solution
We simplify the function before differentiating:
f ( x ) = ( x + 1 x ) 10 = f(x)={(\sqrt{x}+\frac{1}{\sqrt{x}})}^{10}= f ( x ) = ( x + x 1 ) 1 0 =
= ( x + 1 x ) 10 ={(\frac{x+1}{\sqrt{x}})}^{10} = ( x x + 1 ) 1 0
Using Derivative formulas Derivative Rules
f ′ ( x ) = 10 ( x + 1 x ) 9 ⋅ x − ( x + 1 ) ⋅ 1 2 x x = f'(x)=10{(\frac{x+1}{\sqrt{x}})}^9\cdot \frac{\sqrt{x}-(x+1)\cdot\frac{1}{2\sqrt{x}}}{x}= f ′ ( x ) = 1 0 ( x x + 1 ) 9 ⋅ x x − ( x + 1 ) ⋅ 2 x 1 =
One can simplify the derivative:
= 10 ⋅ ( x + 1 ) 9 ( x ) 9 ⋅ x − x + 1 2 x x = =10\cdot\frac{{(x+1)}^9}{{(\sqrt{x})}^9}\cdot \frac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}= = 1 0 ⋅ ( x ) 9 ( x + 1 ) 9 ⋅ x x − 2 x x + 1 =
= 10 ⋅ ( x + 1 ) 9 ( x ) 9 ⋅ 2 x − x − 1 2 x x = =10\cdot\frac{{(x+1)}^9}{{(\sqrt{x})}^9}\cdot\frac{2x-x-1}{2x\sqrt{x}}= = 1 0 ⋅ ( x ) 9 ( x + 1 ) 9 ⋅ 2 x x 2 x − x − 1 =
= 5 ( x + 1 ) 9 ( x − 1 ) x 9 2 ⋅ x ⋅ x 1 2 = =\frac{5{(x+1)}^9(x-1)}{x^{\frac{9}{2}}\cdot x\cdot x^{\frac{1}{2}}}= = x 2 9 ⋅ x ⋅ x 2 1 5 ( x + 1 ) 9 ( x − 1 ) =
= 5 ( x + 1 ) 9 ( x − 1 ) x 6 =\frac{5{(x+1)}^9(x-1)}{x^6} = x 6 5 ( x + 1 ) 9 ( x − 1 )
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