`y = ln(5x + 1)`
The gradient of the above curve at any point is given by the first derivative.
`y=ln(5x+1)`
`dy/dx=[1/(5x+1)] xx 5 = 5/(5x+1)`
Gradient at x=4 is given by;
`(dy/dx)_(x=4) = [5/(5 xx 4 +1)] = 5/21`
So the gradient of curve at x=4 is 5/21
Further Reading
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