Question

Suppose f(0)=2, f’(0)=7, f(1)=5, f’(1)=8, g(1)=1, g’(1)=-9 and H(x)=f(ln(x))+ln(g(x)). Find dH/dx, x=1
What is H’(1)?

Answer 1

Q96195

Solution:

$d/dx(f(x)+g(x))=f'(x)+g'(x)$ -------(1)

And for a composite function f(g(x)), its's derivative is defined as:

$df/dx=f'(g(x))*g'(x)$ -----(2)

Moreover, $d/dx(ln(x))=1/x$ -----(3)

Also, given:

$g(1)=1$ ----(4)

$g'(1)= -9$ ----(5)

$f'(0)=7$ ----(6)

Now, $H(x)=f(ln(x))+ln(g(x))$

thus, using (1), (2) and (3),

$H'(x)=$ $dH/dx=f'(ln(x))/x +g'(x)/g(x)$

hence, $H'(1) = f'(ln(1))/1 +g'(1)/g(1)$ (since, $ln(1) = 0$ )

$=f'(0) + g'(1)/g(1)$

$= 7 + (-9/1)$ (using (4), (5) and (6))

$= 7-9$

= -2 (Answer)