what are deriv's of f(x) = x^3 tan x, g(x) = sin(e^(3x+2)), h(x) = (x^2+1)/(cos x)

[abel muroi]

f(x) = x³ tan x

g(x) = sin(e^{3x + 2})

h(x) = (x² + 1)/(cos x)



For the first one, i used the product rule for derivatives..

(3x²)(tan x) + (sec²x)(x³)

3x² tan x + x³ sec²x (this is my answer)



For the second one, i used the chain rule for derivatives..

f'(g(x)) * g'(x)

f(u) = sin u (the derivative of sin u is cos u)

g(x) = e^{3x + 2} (the derivative of e^{3x + 2} is 3x + 2 e^{3x + 2} .....I think....)

f'(g(x)) * g'(x) = (3x + 2 e^{3x + 2})(cos e^{3x + 2}) (this is my answer)



For the third one, i used the quotient rule for derivatives

((2x)(cos x) - (-sin x)(x²+1))/(cos x)²

(2x cos x + x²sin x + sin x)/(cos x²) (this is my answer)



Can anyone tell me if i made a mistake somewhere? did i get the correct derivatives for the functions?

 

[abel muroi]

If no one responds. Then i will assume I got the correct derivatives..

 

[ksdhart]

The first and third problem's derivatives are correct, but the second problem is not. Your basic process is okay, but you'll actually need to apply the chain rule twice. Try using the chain rule to evaluate the derivative of your g(x)=e^(3x+2) and you should see where you went wrong.

 

[abel muroi]

The first and third problem's derivatives are correct, but the second problem is not. Your basic process is okay, but you'll actually need to apply the chain rule twice. Try using the chain rule to evaluate the derivative of your g(x)=e^(3x+2) and you should see where you went wrong.

AH I get it now. I was supposed to apply the chain rule to e^3x+2 as well

So the correct answer for the second one is cos(e^3x+2) * 3 * e^3x+2

 

[lookagain]

h(x) = (x² + 1)/(cos x)



For the third one, i used the quotient rule for derivatives

((2x)(cos x) - (-sin x)(x²+1))/(cos x)²

(2x cos x + x²sin x + sin x)/(cos x²) (this is my answer)

That answer is also not correct. You have the x squared. It's not.
Cosine of x is squared.


(2xcos x + x²sin x + sin x)/(cos x)² \(\displaystyle \ \ \ \) and

(2xcos x + x²sin x + sin x)/cos²x


are a couple of the correct ways of showing it.

 

[endrits079]

d

derivavetive if sin u is cos u * derivativd of u