Question

We consider f a function defined on R by f(x) = sin(√x), (x >= 0).1) Determine the first derivative of f, f'(x).2) Establish the relation 4xf'(x)² + f(x)² = 1

Answer 1

First, let's calculate the derivative:

$f^{\prime}(x)=\frac{d}{dx}(\sin \sqrt[]{x})$

Here we have a compounded function, so we need to apply the chain rule:

$f^{\prime}(x)=\cos \sqrt[]{x}\cdot\frac{d}{dx}(\sqrt[]{x})$

Finally, we obtain:

$f^{\prime}(x)=\cos \sqrt[]{x}\cdot(\frac{-1}{2\cdot\sqrt[]{x}})$$f^{\prime}(x)=-\frac{\cos \sqrt[]{x}}{2\cdot\sqrt[]{x}}$

Now, let's establish the relationship:

$4x\cdot f^{\prime}(x)^2+f(x)^2=1$

replacing the expressions of the functions:

$4x\cdot(\frac{\cos\sqrt[]{x}}{2\cdot\sqrt[]{x}})^2+\sin ^2\sqrt[]{x}=1$$4x\cdot\frac{\cos^2\sqrt[]{x}}{2^2\cdot\sqrt[]{x}^2}+\sin ^2\sqrt[]{x}=1$$4x\cdot\frac{\cos^2\sqrt[]{x}}{4x}+\sin ^2\sqrt[]{x}=1$

We can cancel the terms 4x in the first term of the expression:

$\cos ^2\sqrt[]{x}+\sin ^2\sqrt[]{x}=1$

From the Pythagorean identity, we know that the sum of cosine squared, plus sine squared of any angle is always equal to 1:

$\sin ^2\theta+\cos ^2\theta=1$

Then, we confirm that the relationship established in 2) is correct:

$\begin{gathered} \cos ^2\sqrt[]{x}+\sin ^2\sqrt[]{x}=1 \\ \\ 1=1 \end{gathered}$