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How do you implicitly differentiate #6=ylny-x#?

Answer 1

Isabella Ackerman

#(dy)/(dx)=1/(lny+1)#

differentiate both side wrt #x#

#d/(dx)(6)=d/(dx)(ylny)-d/(dx)(x)#

points to note:

1) When differentiating #f(y)# wrt #""x"" # differentiate wrt #y# and then multiply by #(dy)/(dx)#

for the first term on RHS the product rule needs to be employed.

#0=(dy)/(dx)lny+cancel(yxx1/y)xx(dy)/(dx)-1#

#0=(dy)/(dx)lny+(dy)/(dx)-1#

Now rearrange for #(dy)/(dx)#

#0=(lny+1)(dy)/(dx)-1#

#(dy)/(dx)=1/(lny+1)#

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Answer 2

Christopher Agresta

#dy/dx=1/(1+lny)#

differentiate #color(blue)"implicitly with respect to x"#

To differentiate #ylny" use " color(blue)"product rule"#

#0=(yxx1/y.dy/dx+lnyxx1.dy/dx)-1#

#rArrdy/dx(1+lny)=1#

#rArrdy/dx=1/(1+lny)#

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Answer 3

Charles Arocha

To implicitly differentiate (6 = y \ln y - x), differentiate each term with respect to (x) and then solve for (\frac{dy}{dx}).

[ \frac{d}{dx}(6) = \frac{d}{dx}(y \ln y) - \frac{d}{dx}(x) ] [ 0 = \frac{dy}{dx} \ln y + y \frac{d}{dx}(\ln y) - 1 ] [ 0 = \frac{dy}{dx} \ln y + \frac{y}{y} \frac{dy}{dx} - 1 ] [ 0 = \frac{dy}{dx} \ln y + \frac{dy}{dx} - 1 ] [ \frac{dy}{dx}(1 + \ln y) = 1 ] [ \frac{dy}{dx} = \frac{1}{1 + \ln y} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.