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Differential dy problem
- Thread starter torque
- Start date
D
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torque said:How do I find the differential dy to the function y = x(sqrt(1-x^2)) ?
I got it down to y = x(1-x^2) ^ 1/2 but I'm stuck from there. Am I supposed to use the chain rule here for the product rule?
\(\displaystyle y \ = \ x * (1 - x^2)^{\frac{1}{2}}\)
\(\displaystyle \frac{dy}{dx} \ = \ 1 * (1 - x^2)^{\frac{1}{2}} + x * \frac{1}{2} * (1 - x^2)^{-\frac{1}{2}} * (-2 * x)\)
Now simplify the above......
Use proper grouping symbols. They mean everything. You have \(\displaystyle 1-\frac{2x^{2}}{\sqrt{1-x^{2}}}\) written, when I think you mean
\(\displaystyle \frac{1-2x^{2}}{\sqrt{1-x^{2}}}\)
You're not wrong. It's the same thing. The book found a common denominator and put it in a different form
\(\displaystyle \sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}}=\frac{1-2x^{2}}{\sqrt{1-x^{2}}}\)
\(\displaystyle \frac{1-2x^{2}}{\sqrt{1-x^{2}}}\)
You're not wrong. It's the same thing. The book found a common denominator and put it in a different form
\(\displaystyle \sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}}=\frac{1-2x^{2}}{\sqrt{1-x^{2}}}\)