chain rule Suppose u and v are differentiable functions of x. Use the given valu

Question

chain rule

Suppose u and v are differentiable functions of x. Use the given values of the functions and their derivatives to find the value of the indicated derivative. 1) (uv) at x = 1 dx A) 36 B)-48 C)-32 D 48 2) 30 10 C)-49 49 (2u-Av) at x = 1 A)-26 B) -18 C) 38 D) 6 write the function in the form y = f(u) and u = g(x). Then find dy/dx as a function of x. 4) y (2x +83 4) dx C) y = u3, u = 2x + 8;dy=3(2x + 8)2 5)y=cos6 x A) y = u6; u cos x; dx = 6 cos5 x sin x B) y _ cos u: u=x6. dx=-6,5 sin(x6) dx y=cos u; u = x Suppose that the functions f and g and their derivatives with respect to x have the following x. Find the derivative with respect to x of the given combination at the given value of x. values at the given values of 6) 3 1 16 8 6) 4 -3 3 5 -6 f(g(x)), x = 4 x(x) g(x) f'(x) g'(x) 4 3 3 5 5 7)1-4 6 5 7)

Explanation / Answer

multiple questions posted.please post each question seperately

10

chain rule: d/dx(uv)=u'v +uv'

at x =1

d/dx(uv)=u'(1)v(1) +u(1)v'(1)

d/dx(uv)=(-6)(7) +(3)(-2)

d/dx(uv)=-42-6

d/dx(uv)= - 48

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2)

quotient rule:d/dx(u/v)=(u'v -uv')/v3

at x =1

d/dx(u/v)=[u'(1)v(1) -u(1)v'(1)]/(v(1))2

d/dx(u/v)=[(-6)(7) -(4)(-3)]/(7)2

d/dx(u/v)=[-42 +12]/49

d/dx(u/v)=-30/49

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3)

d/dx(2u-4v)=2u' -4v'

at x=1

d/dx(2u-4v)=2u'(1) -4v'(1)

d/dx(2u-4v)=(2*-5)-(4*-4)

d/dx(2u-4v)=-10+16

d/dx(2u-4v)=6

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4)

y=u3, u=2x+8 ,

dy/dx=3u2u'

dy/dx=3(2x+8)2(2*1 +0)

dy/dx=6(2x+8)2

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5)

y=cos6x

y=u6, u=cosx ,

dy/dx= 6u5u'

dy/dx= 6(cosx)5(-sinx)

dy/dx= -6cos5x sinx

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