Results 1 to 5 of 5

Math Help - Calc Derivative Questions help plz

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    5

    Calc Derivative Questions help plz

    If anyone could lend some help on these that would be great!

    Find the derivative:

    1. f(x)=sinx-2tanx

    2. y=tan(sinx^2)

    3. f(x)= tan^-1 sq.root(1-x)

    4. y=(secx+tanx)

    5. Let f(x) = 7sin(x+pi)+cos2x
    a. compute f'(x), f''(x), f'''(x), f''''(x)
    b. compute f*13th derivative*(0)
    Last edited by Dr. Noobles; November 25th 2006 at 03:55 PM. Reason: mistype
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Dr. Noobles View Post
    If anyone could lend some help on these that would be great!

    Find the derivative:

    1. f(x)=sinx-2tanx
    Derivative of sine is cosine.
    Derivative of tangent is,
    \sec^2 x
    Thus,
    f'(x)=\cos x-2\sec^2x
    2. y=tan(sinx^2)
    This is a double chain rule,
    y=\tan v
    v=\sin u
    u=x^2
    Thus,
    \frac{dy}{dv}=\sec^2 v

    \frac{dv}{du}=\cos u

    \frac{du}{dx}=2x
    Thus,
    \frac{dy}{dx}=\frac{dy}{dv}\cdot \frac{dv}{du}\cdot \frac{du}{dx}
    Thus,
    \frac{dy}{dx}=\sec^2 v\cos u (2x)
    Thus, substitute,
    \frac{dy}{dx}=2x\sec^2 (\sin u) \cos x^2
    Thus, subsitute again,
    \frac{dy}{dx}=2x\sec^2 (\sin x^2)\cos x^2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2006
    Posts
    5
    thanks ThePerfectHacker

    Anyone have an idea on the others?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by Dr. Noobles View Post
    If anyone could lend some help on these ...
    Find the derivative:
    ...
    3. f(x)= tan^-1 sq.root(1-x)

    4. y=(secx+tanx)
    ...
    Hello,

    to 3. You have to use the chain rule many times:

    \frac{df}{dx}=(-1)\cdot \frac{1}{\tan^2(\sqrt{1-x})} \cdot \frac{1}{\cos^2(\sqrt{1-x})} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-x}} \cdot (-1)
    Split the tan^2 into a fraction of (sin^2)/(cos^2) and cancel the cos^2. You'll get:
    \frac{df}{dx}=\frac{1}{2\sqrt{1-x} \cdot \sin^2(\sqrt{1-x})}

    to 4. I'll use: y=(\sec(x)+\tan(x))=\frac{1}{\cos(x)}+\tan(x)

    \frac{df}{dx}=(-1)\cdot \frac{1}{\cos^2(x)} \cdot (-\sin(x))+\frac{1}{\cos^2(x)}
    Simplify to:
    \frac{df}{dx}=\frac{1+\sin(x)}{\cos^2(x)}

    EB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Dr. Noobles View Post
    3. f(x)= tan^-1 sq.root(1-x)
    Is this
    f(x)= \frac{1}{tan \left ( \sqrt{1-x} \right ) }

    or

    f(x) = tan^{-1} \left (\sqrt{1-x} \right ) where tan^{-1} is the inverse tangent function.

    Earboth did the first one, so in case you need it, I'll do the other.

    f(x) = tan^{-1} \left (\sqrt{1-x} \right )

    This is done by the chain rule just like the others. I'll note that:
    \frac{d}{dx}tan^{-1}(x) = \frac{1}{x^2+1}

    f'(x) = \frac{1}{ (\sqrt{1-x})^2 + 1} \cdot \left ( \frac{1}{2} \cdot \frac{1}{\sqrt{1 - x} } \right ) \cdot (-1)

    f'(x) = - \frac{1}{2 (2 - x) \sqrt{1-x} }

    -Dan
    Last edited by topsquark; November 26th 2006 at 11:32 AM. Reason: LaTeX error
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2 Calc questions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 5th 2010, 05:04 PM
  2. 3 Questions from Calc 3
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 31st 2008, 12:27 PM
  3. Calc questions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 20th 2008, 06:05 PM
  4. few calc/alg questions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 12th 2006, 01:21 PM
  5. Two questions from calc II
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 26th 2005, 06:05 PM

Search Tags


/mathhelpforum @mathhelpforum