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.

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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