lightningzmath
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How would I derivative of radical of √2-x, √3 sinx + cosx. Please help! Thx!
need urgent help in solving some derivative problems
- Thread starter lightningzmath
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Can't really translate the problem statement. Is it these two?
1) \(\displaystyle \dfrac{d}{dx}\sqrt{2-x}\)
2) \(\displaystyle \dfrac{d}{dx}\sqrt{3}\cdot\sin(x) + \cos(x)\)
Maybe it's these two?
3) \(\displaystyle \dfrac{d}{dx}\sqrt{2}-x\)
4) \(\displaystyle \dfrac{d}{dx}\sqrt{3\cdot\sin(x) + \cos(x)}\)
1) \(\displaystyle \dfrac{d}{dx}\sqrt{2-x}\)
2) \(\displaystyle \dfrac{d}{dx}\sqrt{3}\cdot\sin(x) + \cos(x)\)
Maybe it's these two?
3) \(\displaystyle \dfrac{d}{dx}\sqrt{2}-x\)
4) \(\displaystyle \dfrac{d}{dx}\sqrt{3\cdot\sin(x) + \cos(x)}\)
I have assumed that the expressions are all under the radical sign, by using parentheses. otherwise, you would just have a couple of constants.
sqrt(x) = x^(1/2), so use the power rule. Since (1/2)-1 = -1/2, that will move the original square root to the denominator. Use the chain rule to find what has to go in the numerator.
For instance [in cse we don't get back to check your answers till morning],
\(\displaystyle \dfrac{d}{dx} \sqrt{2 - x} =\dfrac{1}{2}\ \dfrac{(-1)}{\sqrt{2 - x}} = - \dfrac{1}{2\ \sqrt{2 - x}}\)
I expect it is (1) and (4)