Recent content by Torshi

Homework Statement 1.) (MVT) f(x) = 2x^3-6x^2-48x+4 on interval [4,9] By the Mean Value Theorem, we know there exists a c in the open interval (-4,9) such that f'( c) is equal to this mean slope. For this problem, there are two values of c that work. The smaller one is __________ and larger...

I figured it out. Thank you. My main issue was with the exponents in regards to if I had to multiply all of them together which was true.

I didn't mean it to be misleading. My fault. The reason why I said that was because I was at my university math lab and even one of the math instructors said it was correct, but made a tad mistake until I mentioned it. I figured out the answer though.

Homework Statement Find the d/dx (x^2+2x+5)^2 Homework Equations Chain rule The Attempt at a Solution My answer: 2(x^2+2x+5)*(2x+2)*2

I think it simplified down to (x^2+4)^8/3 I multiplied the exponents: 1/3 * 4/1 * 2/1 = 8/3

Homework Statement ([2x+1/4x+3]^2) Homework Equations Exponent and quotient rule The Attempt at a Solution Would this become: 2* (2x+1/4x+3) then do the quotient rule?

Oh alrighty! Yea, when I saw the problem I was unsure at first, but then I thought I should do the chain rule. Wasn't too bad now that I think of it. Just wanted to make sure. Thanks!

Homework Statement ([3√(x^2+4)^4]^2 Homework Equations None needed. Chain rule product rule etc The Attempt at a Solution I stopped at: [((x^2+4)^4)^1/3]^2 So I have 3 exponents. I don't know how to simplify this in order to move on to do the chain rule or whatever rule...

Well the original problem was f(x) = (sin(sin(sin(x)))), but yes I believe I got it right. Thank you.

Another problem" Determine (x,y) location(s) where the graph of y^4 = y^2-x^2 has horizontal tangents I got the answer dy/dx = -2x/4y^3-2y I don't know how to calculate x and y positions? I just found the implicit differentiation

Homework Statement Find the equation of y^2=x(x-3)^2 of tangent line at (3,0) Homework Equations Given above. I think implicit differentiation is involved or no since there is no xy's on the same side? The Attempt at a Solution Anyways... My attempt: 2ydy/dx = x*2(x-3)*1...

Homework Statement Derivative of d/dx (sin(sin(sin(x)))) Homework Equations Chain Rule twice? The Attempt at a Solution d/dx Cos(sin(sin(x)))) * Cos(sin(x)) * Cos(x)

Homework Statement Derivative of sin(x^2cos(x)) Homework Equations Product rule and chain rule The Attempt at a Solution [cos(x^2cos(x)) * (2x)(cos(x)) + (x^2)(-sin(x))] ?

Haha alright it's just I have a hard time differentiating between inside and outside so 2*sin(x) with sin(x) turning into 2*cos(x) looks done, but I guess not. Thank you though!

So the exponent 2 comes down multiplying with sin(x) then sin(x) is considered the inside derivative then you multiply that by cos(x)? I don't understand why it just doesn't become 2cos(x)