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Thus, the tangent has gradient
What is the equation of the tangent to the curve # y=9tanx # at the point where #x=(2pi)/3#?
Find the slope of the tangent to the curve y=2/(x+3) at the point where x=a.
Find the equation of the tangent to the curve defined by y =lnx2 / x at the point where x = 4.The answer is: (1 - ln4)x - 8y + (8ln4 - 4) = 0
find the coordinates of the point where the tangent to the curve y=x^3 +x +2 at the point (1,4) meets the curve again. [ans:-2,-8]pls help me i dont understand the question....
y = (ln x2 ) / x
Please use the definition of a derivative function and not the power function. As you go along, also please explain the algebraic methods used to simplify youranswer (rather just writing what you've gotten after using algebra). I need to brush up on algebra and so this is very helpful to me.I cannot simplify further after Thank you.
find the slope of the tangent to the curve y = 3+4x^2-2x^3 at the point where x=a