Let #y = f(x)# be a twice-differentiable function such that #f(1) = 2# and . What is the value of #(d^2y)/(dx^2)# at # x = 1#?

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Answer 1 · 2017-02-18T22:57:40

#(d^2y)/(dx^2) = 28# when #x=1#

Let #y = f(x)# be a twice-differentiable function such that #f(1) = 2# and . What is the value of #(d^2y)/(dx^2)# at# x = 1#?

We have:

#dy/dx = y^2+3#

We know that #y=2# when #x=1# and so:

#y=2 => dy/dx = 2^2+3=7#

Differentiating the above equation (implicitly) wrt #x# we get:

#(d^2y)/(dx^2) = 2ydy/dx #

And so when #x=1# we have:

#(d^2y)/(dx^2) = 2(2)(7) = 28#

Answer 2 · 2017-02-18T23:09:16

#28#

After two derivations, the terms like #a x + b# disappear. Then if

#(dy)/(dx)=y^2+3->(d^2y)/(dx^2)=2y y'=2y(y^2+3)# but

#y(1)=2# so

#(d^2y)/(dx^2)=2 xx 2(2^2+3)=28#