2 answers

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

Question:

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

Answers

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

Explanation:

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)#

2) 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)#

.

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

Explanation:

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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