Calculus: Find $\lim\limits_{h \to 0}{\frac{f(x+h)-f(x)}{h}}$ for $f(x)=\cos(x^2)$

Question

Find the limit:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Given that $f(x) = \cos(x^2)$

Any help will be appreciated.

Answer 1

This is the definition of the derivative of $f(y)=\cos (y^2)$ at the point $y=x$. So, $$ \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x)=-2x\sin(x^2) $$

Answer 2

The derivative can be found by first principles.

Method: use sum to product identity. $$ \begin{align*} \lim_{h \rightarrow0}\frac{\cos(x+h)^2-\cos x^2}{h} &= \lim_{h \rightarrow0}\frac{-2\sin(x^2+hx+h^2/2)\sin(hx+h^2/2)}{h} \\ &=\lim_{h \rightarrow0}-2\sin(x^2+hx+h^2/2) \frac{\sin(hx+h^2/2)}{hx+h^2/2}\frac{hx+h^2/2}{h} \\ &=\lim_{h \rightarrow0}-2\sin(x^2+hx+h^2/2) \frac{\sin(hx+h^2/2)}{hx+h^2/2}(x+h/2)\\ &=-2\sin(x^2)(1)(x)=-2x\sin(x^2) \end{align*} $$

Answer 3

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Let $f(x) = \cos(x^2)$, then

$$\lim_{h \to 0} \frac{\cos((x+h)^2)-\cos(x^2)}{h}$$ Apply L'Hopital's Rule $$\lim _{h\to 0}\left(\frac{-2\sin \left(\left(x+h\right)^2\right)\left(x+h\right)}{1}\right)$$ $$ \lim _{h\to 0}\left(-2\left(x+h\right)\sin \left(\left(x+h\right)^2\right)\right)$$ $$-2\left(x+0\right)\sin \left(\left(x+0\right)^2\right)$$

$$-2x\sin \left(x^2\right)$$

You can also apply the Chain rule: $f(g(x)) = (f\circ g)'=(f'\circ g)\cdot g'$, where in your case $f(x) = \cos(x)$ and $g(x) = x^2$

$$f(g(x)) = \cos(x^2) = -\sin(x^2)\cdot2x = -2x\sin(x^2)$$