Question

微积分数学问题...高手帮忙...

Assume f is a differentiable function whose graph passes through the point (1,4). If g(x) = f(x^2) and the tangent line to the graph of f at (1,4) is y=3x+1, determine g'(x)????

Answer 1

依题目所示，f(x)过点（1,4）且在点（1,4）处的切线为y=3x+1,
因此有f'(x)=3x+1，两边进行积分得到
f(x)=1.5x^2+x+C
由于f(x)过点(1,4)，将x=1，y=4代入上式，得C=1.5
故f(x)=1.5x^2+x+1.5
g(x)=f(x^2)=1.5（x^2）^2+x^2+1.5=1.5x^4+x^2+1.5
g'(x)=6x^3+2x

Answer 2

f'(x)=3x+1
f(x)=1.5x^2+x+C
(1,4)代入上式
C=1.5
f(x)=1.5x^2+x+1.5
g(x)=f(x^2)=1.5x^4+x^2+1.5
g'(x)=6x^3+2x