Find dy/dx of y=1/x. It may be simpler for you to examine the equation y=x^-1. This equation is the exact same as y=1/x. Therefore just multiply -1 by x and subtract 1 from the exponent giving you -x^-2 or y=-(1/x^2).
You can also do it through quotient rules. Therefore take the derivative of the top 1 which = 0 and multiply that by the bottom X which will give you 0. Then subtract the derivative of the bottom x this equals 1 and multiply it by the top (1). Put this all over the bottom squared. Which leads to -1/x^2.
y=1/x = y=x^-1 = -x^-2 = -(1/x^2)=dy/dx
or
y=1/x = ((0*x)-(1*1)/x^2 = dy/dx=-1/x^2
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The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
the derivative of 1x would be 1 the derivative of x to the power of 1 would be 1. the derivative of x+1 would be 1 the derivative of x-1 would be 1 im not sure what you are asking, but however you put it, it's 1.
x/2 is the same as (1/2) times x. Thus, you can use the formula for a constant factor.
It is negative one divided by 4 multiplied by x to the power of 1.5 -1/(4(x^1.5))
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