D/dx[√(16 x^2)] =1/2(16 x^2)^1/2 1.d/dx[16 x^2] =d/dx[16] d/dx[x^2]/2√(16 x^2) =0 2x/2√(16 x^2) = x/√(16 x^2) d/dx[√(16 x^2)] =1/2(16 x)^1/2 1.d/dx[16 x] ...

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Byju's Answer

Standard X

Mathematics

Number Theory: Interesting Results

Find d/d x[√1...

Question

Find $\frac{d}{d x} [\sqrt{16 - x^{2}}]$

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Solution

$\frac{d}{d x} [\sqrt{16 - x^{2}}] = \frac{1}{2} (16 - x^{2})^{\frac{1}{2} - 1} . \frac{d}{d x} [16 - x^{2}] = \frac{\frac{d}{d x} [16] - \frac{d}{d x} [x^{2}]}{2 \sqrt{16 - x^{2}}} = \frac{0 - 2 x}{2 \sqrt{16 - x^{2}}} = - \frac{x}{\sqrt{16 - x^{2}}} \frac{d}{d x} [\sqrt{16 - x^{2}}] = \frac{1}{2} (16 - x)^{\frac{1}{2} - 1} . \frac{d}{d x} [16 - x] = \frac{\frac{d}{d x} [16] - \frac{d}{d x} [x]}{2 \sqrt{16 - x}} = \frac{0 - 1}{2 \sqrt{16 - x}} = - \frac{1}{2 \sqrt{16 - x}}$

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Find ddx[√16−x2]

ddx[√16−x2]=12(16−x2)12−1.ddx[16−x2]=ddx[16]−ddx[x2]2√16−x2=0−2x2√16−x2=−x√16−x2ddx[√16−x2]=12(16−x)12−1.ddx[16−x]=ddx[16]−ddx[x]2√16−x=0−12√16−x=−12√16−x

Find $\frac{d}{d x} [\sqrt{16 - x^{2}}]$