Recent content by Joyci116

thank you.

Homework Statement a.) f(x)=tan2(x) b.) cos3(x2) c.) (2x-1)/(5x+2) d.) (sqrt(x2-2x))(secx) e.) f(x)=((2x+3)/(x+7))3/2 f.) [sin(x)cos(x)]2 Homework Equations chain rule Product rule Quotient rule Power rule The Attempt at a Solution a.) would you do the power rule...

(x2-x)cosx+sinx(2x-1) X2cosx-xcosx+2xsinx-sinx

Homework Statement f(x)=(x2-x)(sin(x))Homework Equations Chain rule Product ruleThe Attempt at a Solution f'g(x)*g'(x) cosx (2x-x)(cosx) Would you use the product rule after using the power rule and chain rule?

Homework Statement A small object with mass 4.00kg moves counterclockwise with constant speed 4.50m/s in a circle of radius 3.00m centered at the origin. It starts at the point with position vector (3.00i+0j)m. Then it undergoes an angular displacement of 9.00rad. (a.) What it is position...

x=2 [2(-3)-4(0)]/2^2 =-3/2

\frac{x[h'(x)]-h(x)x'}{x^{2}}

Um, h(x)=4, because x=2 so h(2)=4; x=2, so I get 1/2. But you don't understand how you would the the quotient rule using that value. There is the product rule if you rearrange the formula to 1(2)^-1 I'm sorry, I'm a little confused.

16! Thank you very much :)

f'(x)=X^(1/2)g'(x)+g(x)(1/2)x^(-1/2) ?

f'(x)=X^(1/2)g'g(x)+g(x)(1/2)x^(-1/2) ?

SO the derivative would be -3/4?

f'(x)=x^{\frac{1}{2}}7+8(\frac{1}{2}x^{-\frac{1}{2}})

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

The product rule is d/dx [f(x)*g(x)]=f(x)g'(x)+g(x)f'(x)