Differentiation Rules are formulas that allow us to take a complicated derivative and break it up into smaller pieces so we can apply more basic differentiation rules.
The Product Rule
The product rule states that for the functions u and v:
First, notice that this is rational function, so we'll need to use the quotient rule. The derivative of the denominator ( x2+3 ) is simply 2x. In order to use the quotient rule, however, we'll also need to know the derivative of the numerator, which we can't find directly.
In order to find the derivative of the numerator, x2+1, we'll use the chain rule:
dxd(x2+1)=2x2+11⋅dxd(x2+1)=2x2+11⋅2x=x2+1x
Now that we know the derivatives of both the numerator and the denominator, we can proceed to use the quotient rule.
f′(x)=(x2+3)2(x2+3)(x2+1x)−x2+1(2x)
Go through each term, and make sure you understand why it's there. You may want to refer again to the statement of the quotient rule above.
Now we should simplify, so factoring out x2+11, we get:
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differentiation of ex is ex only .
we can do it by log method too.
let y= ex ..........> Eq1
taking log both sides we get
log y = log ex
since log and exponential are inversely propotional logexis equal to x
then by differentiating both sides we get , y1 . dxdy = 1 ( w.k.t dxdlog = x1 )
\ (\frac{dy}{dx}) = y1 . 1 =ex .1 = ex ( from eq 1 ) .
hence differentiated
hope u got your answer
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please differentiate e^x
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d(e^x)/dx=e^x
differentiation of ex is ex only . we can do it by log method too. let y= ex ..........> Eq1 taking log both sides we get log y = log ex since log and exponential are inversely propotional logexis equal to x then by differentiating both sides we get ,
y1 . dxdy = 1 ( w.k.t dxdlog = x1 ) \ (\frac{dy}{dx}) = y1 . 1 =ex .1 = ex ( from eq 1 ) . hence differentiated hope u got your answer
We can use the fact that: ∫x1=Lnx(+c) to solve the following differential equation: dxdy=y ⇒∫y1dy=∫1dx⇒ Lny=x+c ∴ y=ex+c ∴ y=ex⟹dxdy=ex
nicely explained!