tangent qn help (1 Viewer)

chinmayagupta · Apr 6, 2015

can some1 help me out for this questions thanks its kinda urgent

integral95 · Apr 6, 2015

Find dy/dy using implicit differentiation (product rule)

then sub the point (2,1) to obtain the gradient of the tangent

then use the point gradient formula

Ambility · Apr 6, 2015

So the question is asking for a equation of a tangent line. In order to find the equation, we need a point the line passes through and the gradient. We already have the point, we now just need the gradient. In order to find the slope of a tangent at a certain point in a function, we need to find the derivative of that function, but first we must simplify.

$x^6y^6=64\\y=\frac{2}{x}\\\text{Now, we take the derivative by the power rule}\\y=\frac{-2}{x^2}\\\text{Now we evaluate the derivative function at the point }x=2\\\frac{-2}{2^2}=-\frac{1}{2}=m\\\text{Now we know the slope of the tangent at the given point, we just need to substitute it into }y=mx+b\text{, with x equaling zero to find our y-intercept.}\\y=mx+b\\1=-\frac{1}{2}2+b\\b=2\\\text{Therefore, the answer is the third answer, }y=-\frac{1}{2}x+2$

braintic · Apr 6, 2015

Ambility said:
So the question is asking for a equation of a tangent line. In order to find the equation, we need a point the line passes through and the gradient. We already have the point, we now just need the gradient. In order to find the slope of a tangent at a certain point in a function, we need to find the derivative of that function, but first we must simplify.

$x^6y^6=64\\y=\frac{2}{x}\\\text{Now, we take the derivative by the power rule}\\y=\frac{-2}{x^2}\\\text{Now we evaluate the derivative function at the point }x=2\\\frac{-2}{2^2}=-\frac{1}{2}=m\\\text{Now we know the slope of the tangent at the given point, we just need to substitute it into }y=mx+b\text{, with x equaling zero to find our y-intercept.}\\y=mx+b\\1=-\frac{1}{2}2+b\\b=2\\\text{Therefore, the answer is the third answer, }y=-\frac{1}{2}x+2$

y is actually equal to +/- 2/x.
If you're going to do it that way, you need to mention that the given point is on the positive curve and not the negative.

Ambility · Apr 6, 2015

braintic said:
y is actually equal to +/- 2/x.
If you're going to do it that way, you need to mention that the given point is on the positive curve and not the negative.

Fair point.

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tangent qn help (1 Viewer)

chinmayagupta

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integral95

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Ambility

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braintic

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Ambility

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