Question

Find
dy/DX
if 5e^x/3e^x+1
at x=0​

Answer 1

Answer:

1) If f(x) = ex, then f '(x) = ex

2) If f(x) = eg(x), then f '(x) = g'(x).eg(x)

3) If f(x) = ln x, then f '(x) = 1/x   (x > 0)

4) If f(x) = ln g(x), then f '(x) = g'(x)/g(x)    [g(x) > 0]

Sample Problems

Step-by-step explanation:

Answer 2

Concept

Differentiation is the change in variable y due to change in variable x. It is expressed as dy/dx if function is f(x). It is the method of finding derivative of a function. It is often used in finding the difference in values of a function.

Given

Function: f(x)=$5e^{x}/3e^{x}+1$

To find

dy/dx of the function f(x)=$5e^{x} /3e^{x} +1$

Explanation

The given function is f(x)=$5e^{x} /3e^{x} +1$ and we have to find dy/dx and we know that dy/dx of function which is in divison form= (vdu/dx-udv/dx)/$v^{2}$

dy/dx=d/dx* $5e^{x} /3e^{x} +1$

=$(3e^{x}+1 *5e^{x}-5e^{x}*3e^{x})/(3e^{x} +1 )^{2}$

Now multiply the expressions and add the powers of the same base which are in multiplication.

=$15e^{2x}+5e^{x}-15e^{2x} /(3e^{x}+1)^{2}$

=$5e^{x}/(3e^{x} +1)^{2}$

Now put the value of x=0.

=$5e^{0}/(3e^{0}+1)^{2}$

=5/16   [$e^{0} =1$]

Hence the value of dy /dx of function $5e^{x}/3e^{x}+1$ at x=0 is 5/16.

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