calculus of trig (1 Viewer)

[totallybord]

hi
just wondering if someone could tell me how to differentiate sinx^3 + 1
thanks

 

[watatank]

chain rule.

d/dx (sinx)³ + 1

= 3(sinx)²(cosx)

or if you meant

d/dx [sin(x³)]+ 1

that's just 3x²[cos(x³)]

 

[totallybord]

what if it was sin(x^3 + 1)?

 

[pLuvia]

3x²cos(x³+1)

Differentiate the inside then differentiate the whole thing

 

[JasonNg1025]

if you can't remember the rule..

Just remember that

dy/dx = dy/du * du/dx [du's cancel out]

where u is a function of x

so like sin³(x):

u = sin(x)
du/dx = cos(x)

so d/dx(sin³(x)) = d/du(u³) * du/dx, and you would use the value of du/dx from above

and dy/dx = 3u² * cos(x)
= 3sin²(x)cos(x)

 

[totallybord]

thanks for helping!
i actually can't remember the exact question so it's one of the 2 listed below or
sin (x+1)^3...if that was the case how would you do that?
thank you!!

 

[watatank]

thanks for helping!
i actually can't remember the exact question so it's one of the 2 listed below or
sin (x+1)^3...if that was the case how would you do that?
thank you!!

well its the chain rule again. it would be cos(x+1)^3 * 3(x+1)^2

= 3(x+1)².cos(x+1)³

 

[JasonNg1025]

Exactly..

But just to clear things up, part of differentiating sin((x+1)³)

involves differentiating (x+1)³.

This is the chain rule because it's not a plain "x" function like x³;

Except that d/dx(x+1) = 1 so multiplying by 1 is negligible

So be careful... sometimes one differentiation will need 2 applications of the chain rule..

 

[totallybord]

thanks a lot!