Differentiate using the definition of differentiation

[Dynamic]

Hi everyone,

My question is as follows:
Use the definition of the derivative to find f′(x) when f(x) is given by:


The definition I have been provided is:




So far I have used the formula (f(x+h) - f(x))/h to simplify my question down to (6*h*x + 3*h^2 + 2)/h although I am struggling to reduce further. Any help would be greatly appreciated.

 

[Deleted member 4993]

Hi everyone,

My question is as follows:
Use the definition of the derivative to find f′(x) when f(x) is given by:
View attachment 28100

So far I have used the formula (f(x+h) - f(x))/h to simplify my question down to (6*h*x + 3*h^2 + 2)/h although I am struggling to reduce further. Any help would be greatly appreciated.

Your work has mistakes. Please show, in detail (step-by-step)

f(x+h) - f(x) = ?

 

[Harry_the_cat]

Have a close look at where the +2 came from on the top.

 

[pka]

Hi everyone,

My question is as follows:
Use the definition of the derivative to find f′(x) when f(x) is given by:
View attachment 28100

The definition I have been provided is:

View attachment 28101


So far I have used the formula (f(x+h) - f(x))/h to simplify my question down to (6*h*x + 3*h^2 + 2)/h although I am struggling to reduce further. Any help would be greatly appreciated.

This a perfect example to show that one really does need to know basic algebra in order to do calculus.
\(f(x)=3x^2+1\)
\(f(x+h)-f(x)\)
\([3(x+1)^2+1]-[3x^2+1]\)
\([3(x^2+2hx+h^2)+1]-[3x^2+1]\)
\(=~?\)
In all of these we need to end-up with an expression that has an \(h\) in each term. WHY?

 

[Dynamic]

This a perfect example to show that one really does need to know basic algebra in order to do calculus.
\(f(x)=3x^2+1\)
\(f(x+h)-f(x)\)
\([3(x+1)^2+1]-[3x^2+1]\)
\([3(x^2+2hx+h^2)+1]-[3x^2+1]\)
\(=~?\)
In all of these we need to end-up with an expression that has an \(h\) in each term. WHY?

I didn't split the two functions up into brackets and therefore was adding the two 1's together... thanks for the help.

 

[JeffM]

I didn't split the two functions up into brackets and therefore was adding the two 1's together... thanks for the help.

That is a very common error, easy to do.