I'm new here and self-studying calculus at the moment. I'm studying linear approximations, but there is a problem I've been stuck on and do not understand the solution. I understand that the formula for linear approximations is: $$f(a)\approx f(a)+f'(a)(x-a)$$
The linearization problem is as follows: Find the linear approximation near $x=0$ for the function $cos(\frac{\pi}2(1+x)^\frac{3}2)$. The solution offered by the course material is attached.
What I don't understand is why the term $-\sin(\frac{\pi}2)$, which represents the $f'(a)$ term in the linear approximation formula, does not seem to apply the chain rule to find the derivative of $cos(\frac{\pi}2(1+x)^\frac{3}2)$ evaluated at $x=0$.
I would think that because this is a composite function, the derivative should have been $$\frac{-3\pi\sqrt(1+x)\sin(\frac{\pi}2(1+x)^\frac{3}2)}4$$ which when evaluated at $x=0$ would be $$-\frac{3\pi\sin(\frac{\pi}2)}4 = -\frac{3\pi}4$$ Instead, the solution only has the $-\sin(\frac{\pi}2(1+x)^\frac{3}2)$ term, which at $x=0$ is $-\sin(\frac{\pi}2) = -1$.
Could someone please kindly explain? Thank you in advance.
The linear approximation for a function $f(x)$ near $x=k$ is given by, $f(x)\approx f(k)+f'(k)(x-k)$.
Here,
$g(x)=\frac{\pi}{2}(1+x)^\frac{3}{2}$
$f(u)=cos(u)$
Near $x=a$,we have $$g(x)\approx g(a)+g'(a)(x-a)$$ $$\frac{\pi}{2}(1+x)^\frac{3}{2}\approx \frac{\pi}{2}(1+0)^\frac{3}{2}+\frac{3\pi(1+0)^{\frac{3}{2}}}{4}(x-0)=\frac{\pi}{2}+\frac{3 \pi x}{4}$$ $$[x=a=0]$$
Near $u=g(a)$, we have $$f(u)\approx f(g(a))+f'(g(a))(u-g(a))$$ $$\cos(u)\approx \cos(\frac{\pi}{2}(1+0)^\frac{3}{2})-\sin(\frac{\pi}{2}(1+0)^\frac{3}{2})(u-g(a))$$ [ We differentiate $f$ w.r.t $g(x)$ and then evaluate at $x=a$] $$=\cos(\frac{\pi}{2})-\sin(\frac{\pi}{2})(u-\frac{\pi}{2}(1+0)^\frac{3}{2})$$ $$=\cos(\frac{\pi}{2})-\sin(\frac{\pi}{2})(u-\frac{\pi}{2})$$ $$\cos(u)\approx \frac{\pi}{2}-u$$
When $u=g(x)$ $$\cos(u)\approx \frac{\pi}{2}-\frac{\pi}{2}-\frac{3 \pi x}{4}= \frac{-3 \pi x}{4}$$
Alternatively, we can use the below linear approximation for composite functions and do it directly as you hinted in your comment.
$$f(g(x))\approx f(g(a))+f'(g(a))(g(x)-g(a))$$ $$f(g(x))\approx f(g(a))+f'(g(a))(g(a)+g'(a)(x-a)-g(a))$$ $$f(g(x))\approx f(g(a))+f'(g(a))(g'(a)(x-a))$$ $$f(g(x))\approx f(g(a))+f'(g(a))g'(a)(x-a))$$
$$f(g(x))=\cos(\frac{\pi}{2}(1+x)^\frac{3}{2}) \approx \cos(\frac{\pi}{2}(1+0)^\frac{3}{2})+(\frac{-3\pi \sqrt{1+0}\sin(\frac{\pi}{2})(1+0)^\frac{3}{2}}{4})(x-0)=-\frac{-3\pi x}{4}$$