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∙ 14y ago- the derivative with respect to x is 40y - The derivative with respect to Y is 40x
So, since both x and y equal 2, both derivatives yield 40*2 = 80
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∙ 14y agox=2
The derivative of 10x is 10. This is irrespective of the value of x.
Yes, the derivative of xi with respect to x equals i. Is that what you were trying to ask?
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
Following the correct order of operations: derivative of x^2 + 6/2 = derivative of x^2 +3, which equals 2x
x=2
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
The derivative of 10x is 10. This is irrespective of the value of x.
Yes, the derivative of xi with respect to x equals i. Is that what you were trying to ask?
x = 10x, so derivative = 10
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
sec(x)tan(x)
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
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))
D(y)= sin 2x