[tex]y=\sin(x^4)[/tex]
[tex]\implies y'=4x^3\cos(x^4)[/tex]
[tex]\implies y'=12x^2\cos(x^4)-16x^6\sin(x^4)[/tex]
At [tex]x=10[/tex], you have
[tex]y'(10)=4000\cos(10^4)[/tex]
The trick to finding out the sign of this is to figure out between which multiples of [tex]\dfrac\pi2[/tex] the value of [tex]10^4[/tex] lies.
We know that [tex]\cos x>0[/tex] whenever [tex]-\dfrac\pi2+2n\pi<x<\dfrac\pi2+2n\pi[/tex], and that [tex]\cos x<0[/tex] whenever [tex]\dfrac\pi2+2n\pi<x<\dfrac{3\pi}2+2n\pi[/tex], where [tex]n\in\mathbb Z[/tex].
We have
[tex]10^4=\dfrac{k\pi}2\implies k=\dfrac{2\times10^4}\pi\approx6366.2[/tex]
which is to say that [tex]\dfrac{6366\pi}2<10^4<\dfrac{6367\pi}2[/tex], an interval that is equivalent modulo [tex]2\pi[/tex] to the interval [tex]\left(\pi,\dfrac{3\pi}2\right)[/tex].
So what we know is that [tex]10^4[/tex] corresponds to the measure of an angle that lies in the third quadrant, where both cosine and sine are negative.
This means [tex]y'(10)<0[/tex], so [tex]y[/tex] is decreasing when [tex]x=10[/tex].
Now, the second derivative has the value
[tex]y'=12\times10^2\cos(10^4)-16\times10^6\sin(10^4)[/tex]
Both [tex]\cos(10^4)[/tex] and [tex]\sin(10^4)[/tex] are negative, so we're essentially computing the sum of a negative number and a positive number. Given that [tex]\sin x>\cos x[/tex] for [tex]\pi<x<\dfrac{5\pi}4[/tex], and [tex]\cos x>\sin x[/tex] for [tex]\dfrac{5\pi}4<x<\dfrac{3\pi}2[/tex], we can use a similar argument to establish in which half of the third quadrant the angle [tex]10^4[/tex] lies. You'll find that the sine term is much larger, so that the second derivative is positive, which means [tex]y[/tex] is concave up when [tex]x=10[/tex].
