Question

Let f and g be differentiable functions such that f′(0)=3 and g′(0)=7. If h(x)=3f(x)−2g(x)−5cosx−3, what is the value of h′(0) ?

Answer 1

To find h′(0), differentiate h(x), resulting in h′(x) = 3f′(x) − 2g′(x) + 5sinx. Use the given derivatives at x=0, f′(0) = 3 and g′(0) = 7, and calculate h′(0) = −5.

Step-by-step explanation:

To find the value of h′(0), we need to differentiate the function h(x) and then substitute x with 0. Given that h(x) = 3f(x) − 2g(x) − 5cosx − 3, the derivative h′(x) will be the sum of the derivatives of each part of h(x).

The derivative of 3f(x) is 3f′(x),

The derivative of −2g(x) is −2g′(x),

The derivative of −5cosx is 5sinx, since the derivative of cosx is −sinx,

The derivative of a constant, such as −3, is 0.

Therefore, combining these, we get:

h′(x) = 3f′(x) − 2g′(x) + 5sinx.

Now we substitute x with 0 and using the given values f′(0) = 3 and g′(0) = 7:

h′(0) = 3(3) − 2(7) + 5sin(0).

h′(0) = 9 − 14 + 0, since sin(0) is 0.

Thus, the value of h′(0) is −5.