Find the equation of tangent line for derivatives of functions

[Dustobusto]

Homework Statement

Find an equation of the tangent line at the point indicated

f(x) = 5^x2-2x+9 , x = 1

Homework Equations

(d/dx) b^x = ln(b)b^x

General Power Rule which states:

(d/dx) g(x)ⁿ = n(g(x))^n-1 * g'(x)

The Attempt at a Solution

So looking at a previous problem first, we have
s(t) = 3^7t, t = 2

So, we can substitute u for the exponent, and take the derivative of that. We get
(ln 3) * 3^u * (7t)' which equals (7 ln 3)3¹⁴

Since it wants to find the tangent line we have to take it further. The equation to the tangent line is y-f(a) = f'(a)(x-a) and point slope form is y = f'(a)(x-a) + f(a)

It helps me to look at it in this scenario as f'(a) = solution before putting it in slope form, a = the given x value(or t in this case), and f(a) = b^x

Using this first example problem that would mean

y = 7 ln 3 * 3¹⁴(t-2) + 3¹⁴, and the book confirms this.


Now, trying to generalize these principles to the problem I am working on..

I switch the exponent with "u" to simplify the problem. We now have (ln 5)5^u(^x2-2x+9)' and since x = 1 its

(ln 5)5^u * 8 which gives us (8 ln 5)5⁸

Attemping to put this into the equation for a tangent line,

f'(a) = (8 ln 5)5⁸
a = 1
f(a) = b^x = 5⁸

so putting that all together is

y = (8 ln 5)5⁸(x-1) + 5⁸...

or so I thought. The book gives

y = 5⁸

How did I screw this up?

 

[Dustobusto]

Ok, I think I know at least the beginning of where I went wrong.

We now have (ln 5)5u(x²-2x+9)' and since x = 1 its (ln 5)5u * 8 which gives us (8 ln 5)5⁸

The derivative of (x²-2x+9) isn't 8. It's 2x-2. I think this is where I need to start.

 

[Mark44]

Homework Statement

Find an equation of the tangent line at the point indicated

f(x) = 5^x2-2x+9 , x = 1

Homework Equations

(d/dx) b^x = ln(b)b^x

General Power Rule which states:

(d/dx) g(x)ⁿ = n(g(x))^n-1 * g'(x)

The Attempt at a Solution

So looking at a previous problem first, we have
s(t) = 3^7t, t = 2

So, we can substitute u for the exponent, and take the derivative of that. We get
(ln 3) * 3^u * (7t)' which equals (7 ln 3)3¹⁴

Since it wants to find the tangent line we have to take it further. The equation to the tangent line is y-f(a) = f'(a)(x-a) and point slope form is y = f'(a)(x-a) + f(a)

It helps me to look at it in this scenario as f'(a) = solution before putting it in slope form, a = the given x value(or t in this case), and f(a) = b^x

Using this first example problem that would mean

y = 7 ln 3 * 3¹⁴(t-2) + 3¹⁴, and the book confirms this.Now, trying to generalize these principles to the problem I am working on..

I switch the exponent with "u" to simplify the problem. We now have (ln 5)5^u(^x2-2x+9)' and since x = 1 its

(ln 5)5^u * 8 which gives us (8 ln 5)5⁸

Attemping to put this into the equation for a tangent line,

f'(a) = (8 ln 5)5⁸
a = 1
f(a) = b^x = 5⁸

so putting that all together is

y = (8 ln 5)5⁸(x-1) + 5⁸...

or so I thought. The book gives

y = 5⁸

How did I screw this up?

f'(1) = 0, which makes the tangent line horizontal.

If f(x) = 5^{x2- 2x + 9} , then f'(x) = (2x - 2) 5^{x2- 2x + 9}, so f'(1) = 0.

 

[Mark44]

It's a lot simpler to convert exponential functions to one that involves e.

For example, 5 = e^ln(5), so 5^x = (e^ln(5))^x = e^{x ln(5)}.

Differentiation is pretty easy, using the fact that d/dx(e^u) = e^u * du/dx.

 

[Mark44]

Ok, I think I know at least the beginning of where I went wrong.

We now have (ln 5)5u(x²-2x+9)' and since x = 1 its (ln 5)5u * 8 which gives us (8 ln 5)5⁸

The derivative of (x²-2x+9) isn't 8. It's 2x-2. I think this is where I need to start.

Yes. You are combining formulas for f'(x) and f'(a) for some number a. Keep them separate, and replace x by a only at the end.

 

[Dustobusto]

f'(1) = 0, which makes the tangent line horizontal.

If f(x) = 5^{x2- 2x + 9} , then f'(x) = (2x - 2) 5^{x2- 2x + 9}, so f'(1) = 0.

Yes, it makes complete sense now. It ends up

(2x-2)(ln 5)5⁸

I believe you can write that out as

2x ln 5 * 2 ln 5 and, x is 1, so

2 ln 5 - 2 ln 5 = 0

Zero times 5⁸ is zero. Got it.

 

[Mark44]

Yes, it makes complete sense now. It ends up

(2x-2)(ln 5)5⁸

I believe you can right that out as

2x ln 5 * 2 ln 5 and, x is 1, so

2 ln 5 - 2 ln 5 = 0

Zero times 5⁸ is zero. Got it.

There's no need to expand the expression. Since x = 1, we can see immediately that 2x -2 = 0.

BTW, in the above, "right" is wrong but "write" is right.

 

[Dustobusto]

There's no need to expand the expression. Since x = 1, we can see immediately that 2x -2 = 0.

BTW, in the above, "right" is wrong but "write" is right.

Yeah, I caught it just before you posted that lol. Thanks Mark.