Check the solution to "find the derivative of g(x)= x^2-4/x+0.5 by quotient..."

Era Metkari

Answered question

2022-09-09

find the derivative of g(x)= x^2-4/x+0.5 by quotient rule

star233

Skilled2023-05-29Added 403 answers

To find the derivative of the function

g (x) = \frac{x^{2} - 4}{x + 0.5}

, we can use the quotient rule. The quotient rule states that if we have a function of the form

f (x) = \frac{u (x)}{v (x)}

, then its derivative can be calculated as:

f^{'} (x) = \frac{u^{'} (x) \cdot v (x) - u (x) \cdot v^{'} (x)}{v (x)^{2}}

In our case,

u (x) = x^{2} - 4

and

v (x) = x + 0.5

. Now, let's calculate the derivatives of

u (x)

and

v (x)

separately.
Using the power rule, the derivative of

u (x)

with respect to

x

is:

\frac{d u}{d x} = \frac{d}{d x} (x^{2} - 4) = 2 x

Similarly, the derivative of

v (x)

with respect to

x

is:

\frac{d v}{d x} = \frac{d}{d x} (x + 0.5) = 1

Now, we can substitute these derivatives into the quotient rule formula:

g^{'} (x) = \frac{(2 x \cdot (x + 0.5)) - ((x^{2} - 4) \cdot 1)}{(x + 0.5)^{2}}

Simplifying further, we have:

g^{'} (x) = \frac{2 x^{2} + x - (x^{2} - 4)}{(x + 0.5)^{2}}

Combining like terms:

g^{'} (x) = \frac{2 x^{2} + x - x^{2} + 4}{(x + 0.5)^{2}}

g^{'} (x) = \frac{x^{2} + x + 4}{(x + 0.5)^{2}}

Therefore, the derivative of

g (x) = \frac{x^{2} - 4}{x + 0.5}

g^{'} (x) = \frac{x^{2} + x + 4}{(x + 0.5)^{2}}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?