Thread: Find Equation of a Line Tangent

I need help finding the line tangent to the graph of the following:

f(x)=4x^3-2x^2+5x-3

at the point x=6

Please help me with this; it's confusing!
So far I took the derivative of the equation and got:
12x^2-4x+5

Where should I go from here?

Any help is appreciated!

ok, so f'(x)=12x^2-4x+5
Now you want to plug in x=6 to find the slope of the tangent line:
12(6)^2-4(6)+5
=432-24+5=413

Now we're going to find the y coordinate when x=6
f(x)=4x^3-2x^2+5x-3
y=4(6)^3-2(6)^2+5(6)-3=819

so your point is (6,819)

Now use that to find the equation for the tangent line:
y-819=413(x-6)
y-819=413x-2487
y=413x-1659

I'm pretty sure that's right. If anybody sees something I missed feel free to comment.

Awesome! Thank you Climberboy123! That was what I was thinking about doing (substitution 6 into the derived equation), but I wasn't sure.

Thanks again!

I have one last problem:

Find the derivative:
y=sqrt(x)*(x+34)

I know that:
(dy/dx)x=1
and
(dy/dx)34=0
but I am uncertain about (dy/dx)sqrt[x].

So would I have (so far):
y=(dy/dx)sqrt[x]*1
=(dy/dx)sqrt[x]
?

Where do I go from here?
Again, help is appreciated!

You took the derivative of part of a product, which you can't do. The only way I can see to solve it is to use the product rule.

the sqrt(x)=x^(1/2), so the derivative would be (1/2)*x^(-1/2) which =1/(2*sqrt(x))

If you know how to use the product rule for derivatives, I would use that.
It's:
(dy/dx)[f(x)*g(x)]=g'(x)*f(x)+f'(x)*g(x)
I hope that helps. If that doesn't make sense I can work it out.

If you wouldn't mind working it out, Climberboy, I'd appreciate it very much! The only rules I know of so far are the power rule (or is this similar to the product rule?), linear functions, and sums/differences. If there is a way these can be used (in combination) to solve this, please work it out for me. Appreciate the replies!

ok, I just took another look at the problem and you can do it without the product rule.
So first we want to multiply it all out:

f(x)=sqrt(x)*(x+34)=x^(1/2)*(x+34)=x^(1/2)*x+x^(1/2)*34

=x^(3/2)+34x^(1/2)

So, now we can use the power rule to find f'(x)

f'(x)=(3/2)x^(1/2)+(34/2)x^(-1/2)= (3x^(1/2))/2+34/(2x^(1/2))

multiply the first part by [x^(1/2)]/[x^(1/2)] to get a common denominator:

=[(3x^(1/2))/2]*[x^(1/2)]/[x^(1/2)]+34/(2x^(1/2))

=(3x)/(2x^(1/2))+34/(2x^(1/2))

=(3x+34)/(2x^(1/2))

=(3x+34)/(2*sqrt(x))

I hope you can follow that. It's kind of hard to read with all the brackets.

I'll have to look that over tomorrow, as my brain doesn't quite work well late at night

Thanks for the replies, I'm sure it will make sense to me as I look it over!