Differentiate with Respect to X

[matt_crouch]

Homework Statement

Differentiate with respect to X

Homework Equations

1) e^3x + ln2x

2) (5+x^2)^3/2

The Attempt at a Solution

1) isn't it just normal differentiation? so

3e^3x + 1/2x

only thing i wasnt sure about was the differentiation of ln2x


2) second one i let t=(5+x^2) differentiated that then differentiated t^3/2 to get 3/2t^1/2

but this is where i got stuck as i wasnt sure where to go from here

is the answer just

3/2 (5+x^2)^1/2 or is it 3/2(2x)^1/2

 

[kbaumen]

Both are examples of the chain rule.
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$

In 1) you've differentiated incorrectly the term ln(2x). The outer function is y=ln(u) but the inner function is u=2x.
So from the result of differentiating ln(2x) should be (1/2x)*2 = 1/x

In 2) the outer function is y=u^(3/2), but the inner function is u=5+x^2. Can you do it from here?

 

[matt_crouch]

so y=u^(3/2) differentiates to dy/dx= (3/2)(x^2)^(1/2)

?

 

[Dick]

No, no, no. y=(5+x^2)^(3/2), u=5+x^2. So y=u^(3/2), dy/dx=dy/du*du/dx. dy/du=d(u^(3/2))/du=(3/2)*u^(1/2)=(3/2)*(5+x^2)^(1/2). That's the part you've got. But you keep forgetting to multiply by the du/dx part. Look up the chain rule and work some more examples. This is a pretty important point.

 

[HallsofIvy]

By the way ln(2x)= ln(x)+ ln(2). That should be easier to differentiate.

 

[matt_crouch]

ahh ye course iv got it now.. it was just the final step i was missing
cheers.
ye i realized the ln(2x)=ln(x)+ln(2) is lost during differntiation because its a constant right?

just lack of revision I am afraid =]