- #1
the1024b
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How can i find the derivative of a function like this:
f(x) = sqrt( 1 - x² )
f(x) = sqrt( 1 - x² )
mathsn00b said:Hi,
I have a similar problem, I need to differentiate sqrt(x^2 + y^2) in terms of x and y.
Starting this I took the simple step (x^2 + y^2)^(1/2)...
My next step is a guess and I am lost after it...(1/2)(x^2 + y^2)(-1/2)...
Any help would be much appreciated.
mathsn00b said:thanks, would I do this by...
df/dx = 1/2(x^2 + y^2)^(-1/2).2x = x/sqrt(x^2 + y^2) and...
df/dy = 1/2(x^2 + y^2)^(-1/2).2y = y/sqrt(x^2 + y^2) ?
thanks for your help so quickly.
If that is meant to be 4^(5(sqrt(x^5))), then you can easily rewrite this to equal68Pirate said:What if i have a problem similar to these however now its 4/ ^5sqrt(x^5)
A derivative of a square root is the rate of change of a square root function at a specific point. It represents the slope of the tangent line to the square root function at that point.
To find the derivative of a square root, you can use the power rule and chain rule. First, rewrite the square root as an exponent of 1/2. Then, apply the power rule and chain rule to find the derivative.
The derivative of the square root of x is 1/2 times the reciprocal of the square root of x. In other words, it is 1/2 divided by the square root of x.
Sure, let's find the derivative of f(x) = √x at x = 4. First, rewrite the square root as x^(1/2). Then, use the power rule and chain rule to find the derivative: f'(x) = 1/2 * x^(-1/2) * 1 = 1/2√x. Substituting x = 4, we get f'(4) = 1/2√4 = 1/4.
No, the derivative of a square root can be positive, negative, or zero, depending on the value of x. For example, at x = 4, the derivative of √x is positive (1/4), but at x = 0, the derivative is undefined. It is important to consider the domain of the function when determining the sign of the derivative.