Unsure on how to approach sin^4

[helpm3pl3ase]

For d/dx (cos 4x + sin^(4) x)

is it.

-sin(4x)(4) + cos^4 (x)? I was unsure on how to approach sin^4. Thanks.

Also for d/dx (1)/(16-t^2)^(1/4)

Is this correct: (-1/4 (16-t^2)^(-5/4))\((16-t^2)^(1/4))^2??

Thank you.

 

[courtrigrad]

For (1) $$\frac{d}{dx} \cos 4x + \sin^{4} x = -4\sin 4x + 4\sin^{3} x \cos x$$. You have to use both the power rule and the chain rule.

$$\frac{d}{dx} \sin^{4}x, \ u = \sin x$$

$$\frac{d}{du} u^{4} du = 4u^{3} du$$For (2) the first half is right. The second half, you have to use the chain rule.

 

[helpm3pl3ase]

Y use the chain rule isn't the second half the bottom of the quotient rule, so it stays like that?/

 

[courtrigrad]

$$\frac{d}{dt} (16-t^{2})^{-\frac{1}{4}} = -\frac{1}{4}(16-t^{2})^{-\frac{5}{4}}(-2t)$$. So we have used the chain rule.

 

[helpm3pl3ase]

O that goes on the top.. the bottom is correct tho right?

 

[HallsofIvy]

quote= helpm3pl3ase]Also for d/dx (1)/(16-t^2)^(1/4)

Is this correct: (-1/4 (16-t^2)^(-5/4))\((16-t^2)^(1/4))^2??[/quote]
You can use the quotient rule (with the chain rule):
$$\frac{(d/dx(1))(16-t^2)^{1/4}- (1)(d/dx(16-t^2)^{1/4})}{(16-t^2)^{1/2}}$$
$$= -\frac{(1)(1/4)(16-t^2)^{-3/4}(-2t)}{(16-t^2)^{1/2}}$$
$$= \frac{1}{2}\frac{t}{(16-t^2)^{5/4}}$$

But it is much easier to write the function as
[tex](16- t^2)^{-1/4}[/itex]
and use the chain rule directly:
$$d/dx(16- t^2)^{-1/4}= -1/4(16- t^2)^{-5/4}(-2t)$$
$$= \frac{1}{2}\frac{t}{(16-t^2)^{5/4}}$$