The curve C has a equation y=(2x-3)^5; point P (0.5,-32)lies on that curve. Work out the equation to the tangent to C at point P in the form of y=mx+c

Firstly you should work out the first derivative of the equation y= After differentiating the equation, sub the x value of the point P into the first derivative. This should give you the gradient of the equation. After getting the gradient of the tangent, you could use the y and x value of point P and sub it into the equation of y=mx+c to work out the y intercept (c). This would give you the answer.

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Answered by Mohammed R. Maths tutor

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The curve C has the parametric equations x=4t+3 and y+ 4t +8 +5/(2t). Find the value of dy/dx at the point on curve C where t=2.


How do you integrate y = 4x^3 - 5/x^2?


differentiate- X^3- 2X^2+3


Find dy/dx where y=e^(4xtanx)


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