We will use the quotient rule here. express sqrt(2) as 2 1/2.
d/dx[f(x)/g(x) = [g(X)f'(X) - f(x)g'(x)]/(g(x)2
d/dx[t7 * (2 1/2)' - 2 1/2 * (t7)']/(t7)2
= t7 * 0 - 2 1/2 * 7t6/t14
return to sqrt(2)
= 7sqrt(2)t6/t14
============ cleaned up
7sqrt(2)/t8
========
sqrt(x) = x^(1/2) The derivative is (1 / 2) * x^(-1 / 2) = 1 / (2 * x^(1 / 2)) = 1 / (2 * sqrt(x))
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y=(8x).5 + (4x).5 = (2+2sqrt(2))x.5 y'=(1 + sqrt(2))/sqrt(x)
In general, if you're taking the derivative with respect to X, then you take the current power of X, multiply the given quantity by that number and then subtract one from the current power. In this case, that's an overcomplicated way of describing what happens but here's the process: 5x is more fully 5*x^1 So you take the power (1) and multiply it it by the given quantity. This gives you 1*5*x^1 Now you subtract one from the current power giving you 1*5*x^0 which equals 5. So the answer is simply 5 in this case. But what if you were trying to find the derivative of 4x^7? In this case, you would multiply the quantity by 7 (giving you 7*4*x^7) and subtract 1 from the current power giving you a final answer of 28*x^6. This also works for negative powers and square roots. The derivative of sqrt(x) can be found by recognizing that this is equal to x^(1/2). So you multiply everything by 1/2 and subtract one from the power and get 1/2 * x^(-1/2) which equals 1/2 * 1/sqrt(x) = 1/(2*sqrt(x))
4/x can be written as 4x-1 (the power of negative 1 means it is the denominator of the fraction) 4*-1 = -4 Therefore, the derivative is -4x-2
The derivative of sqrt(2) is zero.
Derivative of x = 1, and since sqrt(x) = x^(1/2), derivative of x^(1/2) = (1/2)*(x^(-1/2))Add these two terms together and derivative = 1 + 1/(2*sqrt(x))
sqrt(x) = x^(1/2) The derivative is (1 / 2) * x^(-1 / 2) = 1 / (2 * x^(1 / 2)) = 1 / (2 * sqrt(x))
3x - 4 sqrt(2)The first derivative with respect to 'x' is 3.
The anti-derivative of sqrt(x) : sqrt(x)=x^(1/2) The anti-derivative is x^(1/2+1) /(1/2+1) = (2/3) x^(3/2) The anti-derivative is 4e^x is 4 e^x ( I hope you meant e to the power x) The anti-derivative of -sin(x) is cos(x) Adding, the anti-derivative is (2/3) x^(3/2) + 4 e^x + cos(x) + C
The formula for the derivative of an inverse (finv)' = 1/(f' o (finv)) allows you get a formula for the derivative of the inverse of any function that you already know the derivative of. For example: What is the derivative of sqrt(x)? You could figure this out using the definition of the derivative, but it is complicated. You already know that the derivative of x2 is 2x. So let f = x2; finv = sqrt(x), f' = 2x. This gives: (sqrt(x))' = 1/(2 sqrt(x)). Now you have derived a "square root rule" with almost no work.
sqrt(X) is also X^1/2 use power rule 1/2X^-1/2 ( first derivative ) -1/4X^-3/2 ( second derivative ) and so on
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
Take the derivative of (3x+1)1/2First use the power rule on the entire function, then multiply it by the derivative of the inside:(1/2)(3x-1)-1/2(3) = (3/2)(3x-1)-1/2which can also be written as3/(2sqrt(3x-1))(sqrt stands for square root)
By using a form of 1, sqrt(2)/sqrt(2),and multiplying. 15/sqrt(2) * sqrt(2)/sqrt(2) = 15sqrt(2)/2 -------------------
Because questions can't use punctuation the question is ambiguous. Two possible interpretations of the question, with answers are given below:sqrt(3) + sqrt(6) / sqrt(2) = sqrt(3) + sqrt(6/2) = sqrt(3) + sqrt(3) = 2*sqrt(3)The more likely question is:[sqrt(3) + sqrt(6)]/sqrt(2)Multiply numberator and denominator by sqrt(2) to give[sqrt(2)*sqrt(3) + sqrt(2)*sqrt(6)]/[sqrt(2)*sqrt(2)]=[sqrt(2)*sqrt(3) + 2*sqrt(3)]/2 since sqrt(2)*sqrt(2) = 2= sqrt(3)*[sqrt(2)+1]/2AnswerSquare root three plus square root six is square root 9 over square root two. But nine can be factored out to a perfect three so you would have 3 over square root two.
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