Question

If f(x) = ex and g(x) = loge x, then (gof)’ (x) is ?​

Answer 1

Answer:

Step-by-step The use of the function  logx  is problematic in itself - in early mathematics, it denotes the logarithm with base  10 , in more advanced mathematics, one with base  e , and in computer science, one with base  2 .

Because I am unaware of the OP’s exact mathematical level, I will assume that  logx  denotes  logbx  and solve the problem for a general base, leaving it up to you to plug in whatever base for which you are interested in solving.

To compose our two functions, we simply substitute  g(x)  for the  x  in  f(x) :

(f∘g)(x)=f(g(x))=elogbx

We can further simplify this by recalling the change of base formula  logab≡lognalognb  for any  n .

=elnxlnb=(elnx)1lnb

A final thing to note is domain: because in the original logarithm function,  x  must be positive, we have a restricted domain for  x .

(f∘g)(x)=x1lnb, x∈(0,∞)

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Answer 2

(gof)’ (x) is equal to 1.

Step-by-step explanation:

According to the given information,

The given functions are,

f(x) = $e^{x}$ and g(x) = $log_{e}x$.

Now, g(f(x)) is equal to (gof)(x) = $log_{e} e^{x}$

Now, by the properties of logarithm,

(gof)(x) = $log_{e} e^{x}$ is nothing but equal to x.

Thus, the value of  (gof)(x) is equal to x.

Now, we need to find the value of  (gof)’ (x), that is, we need to find the differentiation of  (gof) (x), that is we need to find  $\frac{d}{dx}$(gof)(x).

Now,  (gof)’ (x)

= $\frac{d}{dx}$(gof)(x)

=$\frac{d}{dx}$(x)

=1

Now, this is because x has a power of 1 and $\frac{d}{dx}(x^{n} ) = nx^{n-1}$.

So, when we put n = 1, we get, the result as 1.

Thus, (gof)’ (x) is equal to 1.

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