y = x + ex
then dy/dx = 1 + ex
and d2y/dx2 = ex
2x
well if you're finding the derivative with respect to x, it would be -tx^(-t-1)
d/dx(X^4) = 4X^3 ( first derivative ) d/dx(4X^3) = 12X^2 ( second derivative )
d/dx(2x) = 2 simple power rule
The derivative is 2x based on the power rule. Multiply the power by the coefficient of x then drop the power by one.
2x is the first derivative of x2.
2x
well if you're finding the derivative with respect to x, it would be -tx^(-t-1)
d/dx(X^4) = 4X^3 ( first derivative ) d/dx(4X^3) = 12X^2 ( second derivative )
d/dx(2x) = 2 simple power rule
eight to the second power equal to 64
(x3)'=3x2(x3)''=(3x2)'=6x
With respect to x, the derivative would be:1*Y^3 = Y^3With respect to Y the derivative would be:3*xy^2 - 3In general: the derivative of a variable is defined as: nax^n-1Where n represents the power, a represents the factor and x represents the variable
24u to the second power. Differentiate 40u to the fifth power which is 200u to the fourth power and 5u to the second power which is 10u. Subtract 400u to the sixth power from 1000u to the sixth power which is 600u to the sixth power. Then square 5u to the second power which is 25u to the fourth power. Finally, divide 600u to the sixth power by 25u to the fourth power. The solution is 24u to the second power. Another method is simplifying it to 8u cubed (to the third power) and taking the power rule. Take 3 times 8u which is 24u and subtract 1 from 3 in exponent which is 2. The answer is 24u to the second power.
The derivative is 2x based on the power rule. Multiply the power by the coefficient of x then drop the power by one.
No.
The derivative of x is not x. X is the same as x^1, so you use the power rule which decreases the power by one and brings the exponent down, giving 1x^0, which is equal to 1.