Given that f is a differentiable function with f(3)=1, f′(3)=2. Find h′(x) and h′(3) where h(x)=(x^2 f(x))/(6-5f(x))?

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Answer 1 · 2018-02-15T17:18:00

#h'(x)=((2xf(x)+x^2f'(x))(6-5f(x))-x^2f(x)(-5f'(x)))/(6-5f(x))^2# and #h'(3)=114#

As #h(x)=(x^2 f(x))/(6-5f(x))#

#h'(x)=((2xf(x)+x^2f'(x))(6-5f(x))-x^2f(x)(-5f'(x)))/(6-5f(x))^2#

and #h'(x)=((6f(3)+9f'(3))(6-5f(3))-9f(3)*(-5f'(3)))/(6-5f(3))^2#

= #((6+9*2)(6-5)-9(-5*2))/(6-5)^2#

= #(24+90)#

= #114#

Answer 2 · 2018-02-15T18:15:44

# h'(x)=[2x{3xf'(x)+6f(x)-5(f(x))^2}]/{6-5f(x)}^2#.

# h'(3)=114#.

Let, #h(x)=(u(x))/(v(x))#, where,

#u(x)=x^2f(x), and, v(x)=6-5f(x)#.

Then, by the Quotient Rule, we have,

#h'(x)={v(x)u'(x)-u(x)v'(x)}/(v(x))^2..................................(star)#.

Now, #u(x)=x^2f(x)#, then, using the Product Rule,

#u'(x)=x^2f'(x)+f(x)(x^2)'#,

#:. u'(x)=x^2f'(x)+2xf(x).........................................(star_1)#.

#v(x)=6-5f(x) rArr v'(x)=(6)'-(5f(x))'#,

#:.v'(x)=-5f'(x)......................................................(star_2)#.

Utilising #(star_1) and (star_2)" in "(star)#, we have,

#h'(x)=[(6-5f(x)){x^2f'(x)+2xf(x)}-x^2f(x){-5f'(x)}]/{6-5f(x)}^2#,

#rArr h'(x)={6x^2f'(x)+12xf(x)-10x(f(x))^2}/{6-5f(x)}^2, i.e., #

# h'(x)=[2x{3xf'(x)+6f(x)-5(f(x))^2}]/{6-5f(x)}^2#.

Hence, #h'(3)=[6{9f'(3)+6f(3)-5(f(3))^2}]/{6-5f(3)}^2#,

#=[6{9*2+6*1-5(1)^2}]/{6-5*1}^2#,

#rArr h'(3)=114#.

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