Differential coefficient of the following functions.
Problem From Differentiation.
Answers
1
2
3
4
5
6
7
-1
1
2
3
-1
-2
-3
-4
x
y
1
2
slope = 1/2
Open image in a new page
The graph of \displaystyle{y}= \ln{{\left({x}\right)}}y=ln(x) showing the tangent at \displaystyle{x}={2}.x=2.
The slope of the tangent of y = ln x at \displaystyle{x}={2}x=2 is \displaystyle\frac{1}{{2}}
2
1
. (We can observe this from the graph, by looking at the ratio rise/run).
If y = ln x,
\displaystyle{x}x 1 2 3 4 5
slope of graph \displaystyle{1}1 \displaystyle\frac{1}{{2}}
2
1
\displaystyle\frac{1}{{3}}
3
1
\displaystyle\frac{1}{{4}}
4
1
\displaystyle\frac{1}{{5}}
5
1
\displaystyle\frac{1}{{x}}
x
1
\displaystyle{1}1 \displaystyle\frac{1}{{2}}
2
1
\displaystyle\frac{1}{{3}}
3
1
\displaystyle\frac{1}{{4}}
4
1
\displaystyle\frac{1}{{5}}
5
1
We see that the slope of the graph for each value of x is equal to \displaystyle\frac{1}{{x}}
x
1
. This works for any positive value of x (we cannot have the logarithm of a negative number, of course).
If we did many more examples, we could conclude that the derivative of the logarithm function y = ln x is
\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{1}{{x}}
dx
dy
=
x
1
Note 1: Actually, this result comes from first principles.
Note 2: We are using logarithms with base e. If you need a reminder about log functions, check out Log base e from before.
Answer:
plzzz give me brainliest ans and plzzzz follow me please
