### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Derivaties <br> lecture - 2 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ $\rightarrow$ <span style = "color:blue"> Derivaties lecture - 2 </span> $\rightarrow$ Chain Rule for derivatives $\rightarrow$ Chain Rule for derivatives </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Chain Rule for derivatives </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Recall: Composition of two functions. - Notation: $(f \circ g)(x)=f(g(x))$ - Suppose $g(x)$ in differentiable at $x=a$ and $f(x)$ is differentiable at $x=g(a)$. - Then is $(f \circ g)(x)$ differentiable at $x=a$ ? - If so, $(f \circ g)^{\prime}(a)=$ ? </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ Derivaties lecture - 2 $\rightarrow$ <span style = "color:blue"> Chain Rule for derivatives </span> $\rightarrow$ Chain Rule for derivatives $\rightarrow$ Chain Rule for derivatives </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Chain Rule for derivatives </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Let $k(x)=(f \circ g)(x)=f(g(x))$ - To check whether $k(x)$ is differentiable at $x=a$, we need to $100 k$ at the limit $\lim _{h \rightarrow 0} \frac{k(a+h)-k(a)}{h}$ - i.e. $\lim _{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{h}$ - If $g(a+h)-g(a) \neq 0$ for all small $h \neq 0$ then </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Derivaties lecture - 2 $\rightarrow$ Chain Rule for derivatives $\rightarrow$ <span style = "color:blue"> Chain Rule for derivatives </span> $\rightarrow$ Chain Rule for derivatives $\rightarrow$ Theorem </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Chain Rule for derivatives </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - $\begin{gathered}\frac{f(g(a+h))-f(g(a))}{h}=\frac{f(g(a+h))-f(g(a))}{g(a+h)-g(a)} \cdot \frac{g(a+h)-g(h)}{h} \\ \end{gathered}$ - $\text { if } g(a+h)-g(a) \neq 0$ - we are given that $g$ is diff. at $a$, so - $\lim _{h \rightarrow 0} \frac{g(a+h)-g(a)}{h}=g^{\prime}(a)$ - Also, $f$ in diff. at $g(a)$, so - $\lim _{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{g(a+h)-g(a)}=f^{\prime}(g(a))$ - because as $h \rightarrow 0, g(a+h) \rightarrow g(a)(\because g$ is continuous at a) - $\therefore \lim _{b \rightarrow g(a)} \frac{f(y)-f(g(a)}{y-g(a)}=f^{\prime}(g(a))$. - So, if $g(a+h) \neq g(a)$ for small $h$, then - $(f \circ g)^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a) \text {. }$ - In fact, the above formula is always true. </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Chain Rule for derivatives $\rightarrow$ Chain Rule for derivatives $\rightarrow$ <span style = "color:blue"> Chain Rule for derivatives </span> $\rightarrow$ Theorem $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Theorem </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - (Chain Rule): Let $f$ \& $g$ be functions such that $f$ is diff. at $g(a)$ and $g$ is diff. at $a$. Then $f \circ g$ is diff. - at $a$ and the derivative is given by - $(f \circ g)^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Chain Rule for derivatives $\rightarrow$ Chain Rule for derivatives $\rightarrow$ <span style = "color:blue"> Theorem </span> $\rightarrow$ Proof $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Proof </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - We need to show that - $\lim _{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{h}=f^{\prime}(g(a)) g^{\prime}(a)$ - Define $\phi(y)= \begin{cases}\frac{f(y)-f(g(a))}{y-g(a)} & \text { if } y \neq g(a) \\\\ f^{\prime}(g(a)) & \text { if } y=g(a)\end{cases}$ - $\begin{aligned}\lim _{y \rightarrow g(a)} \phi(y) & =\lim _{y \rightarrow g(a)} \frac{f(y)-f(g(a))}{y-g(a)} \\\\& =f^{\prime}(g(a))[\because f \text { is ff at } g(a)] \\\\& =\phi(g(a))\end{aligned}$ - $\Rightarrow \phi$ is continuous at $g(a)$. </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Chain Rule for derivatives $\rightarrow$ Theorem $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Proof $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Proof </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Note that $f(y)-f(g(a))=\phi(a)[y-g(a)] \quad \forall y$ - (if $y \neq g(a)$ then $\phi(y)=\frac{f(y)-f(g(a))}{y-g(a)}$ - $\Rightarrow f(y)-f(g(a))= \phi(y)[y-g(a)]$ - if $y=g(a)$, the L.H.S $=0=$ R.H.S. ) - Put $y=g(a+h)$ : - $f(g(a+h))-f(g(a))=\phi(g(a+h))[g(a+h)-g(a)]$ - $\therefore$ if $h \neq 0, \frac{f(g(a+h))-f(g(a))}{h}=\phi(g(a+h))\left[\frac{[g(a+h)-g(a)}{h}\right]$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Theorem $\rightarrow$ Proof $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Proof $\rightarrow$ Remark </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Proof </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - In R.H.S., $\lim _{h \rightarrow 0} \frac{g(a+h)-g(a)}{h}=g^{\prime}(a)$ - and $\quad \lim _{h \rightarrow 0} \phi(g(a+h))=\phi(g(a))[\because \phi$ is cont. at g(a)] - $=f^{\prime}(g(a))$ - $\therefore \lim _{h \rightarrow 0} \frac{(f \circ g)(a+h)-(f \circ g)(a)}{h}=f^{\prime}(g(a)) g^{\prime}(a)$. - ie. $(f \circ g)^{\prime}(a)=f^{\prime}(g(a)) \cdot g^{\prime}(a)$ - This proves the chain rule. </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Proof $\rightarrow$ Proof $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Remark $\rightarrow$ Remark </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Remark </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - $\quad g(a+h)-g(a) \neq 0$ might not hold true for small enough $h$. - e.g. $\quad g(x)= x^2 sin(\frac{1}{x}), \quad x \neq 0$, otherwise $g(x)=0$ - Claim: $g$ is diff. at $0 ~ \\& ~ g^{\prime}(a)=0$ - $\lim _{h \rightarrow 0} (\frac{g(0+h)-g(0)}{h}) =\lim _{h \rightarrow 0} \frac{h^2 \sin \left(\frac{1}{h}\right)-0}{h}$ - $ =\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)=0$ - $\left|\sin \frac{1}{h}\right| \leq 1 \Rightarrow 0 \leq h \sin \left(\frac{1}{h}\right)|\leq| h \mid$ - $\text { By Sandwich thin., } \left.\lim _{h \rightarrow \infty} h \sin \left(\frac{h}{h}\right)=0\right)$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Proof $\rightarrow$ Proof $\rightarrow$ <span style = "color:blue"> Remark </span> $\rightarrow$ Remark $\rightarrow$ Other forms of chain rule </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Remark </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - But, $g\left(\frac{1}{m \pi}\right)=0$ for all m $\in \mathbb{Z}$\{0} - because $\sin (m \pi)=0 \quad \forall m \in \mathbb{Z}$ - For any $\delta>0, \frac{1}{m \pi} \in(-\delta, \delta)$ for some m $\in \mathbb{Z}$. </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Proof $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Remark </span> $\rightarrow$ Other forms of chain rule $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Other forms of chain rule </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Write $y=g(x)$ & $u=f(y)$ - ie. $u=f(g(x))=(f \circ g)(x)$ - The chain rule $(f \circ g)^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x)$ - ie. $\frac{d u}{d x}=\frac{d u}{d y} \cdot \frac{d y}{d x} \quad\left(\begin{array}{l}y=g(x) \Rightarrow \frac{d y}{d x}=g^{\prime}(x) \\ \left.u=f(y) \Rightarrow \frac{d u}{d y}=f^{\prime}(y)\right)\end{array}\right.$ - $u=f(y) ; \quad y=g(x)$ - $\frac{d u}{d x}=\frac{d u}{d y} \cdot \frac{d y}{d x}$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Remark $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Other forms of chain rule </span> $\rightarrow$ Example 1 $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Example 1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Let $f(x)=\left(x^2+1\right)^3$. Find $f^{\prime}(x)$. - $\rightarrow$ Expand $\left(x^2+1\right)^3=x^6+3 x^4+3 x^2+1$ - $\therefore f^{\prime}(x)=6 x^5+12 x^3+6 x $ - $\rightarrow$ Use product rule: $f(x)=\left(x^2+1\right)\left(x^2+1\right)^2$ - $f^{\prime}(x) =2 x\left(x^2+1\right)^2+\left(x^2+1\right) \cdot \frac{d}{d x}\left(x^2+1\right)^2$ - $\frac{d}{d x}\left(x^2+1\right)^2 =2 x\left(x^2+1\right)+\left(x^2+1\right) \cdot 2 x$ - $ =4 x\left(x^2+1\right)$ - $\therefore \quad f^{\prime}(x) =2 x\left(x^2+1\right)^2+4 x\left(x^2+1\right)^2$ - $ =6 x\left(x^2+1\right)^2$ - $ =6 x\left(x^4+2 x^2+1\right)$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Remark $\rightarrow$ Other forms of chain rule $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 1 $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Example 1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Using chain rule: - $f(x)=\left(x^2+1\right)^3=(g(x))^3 ; g(x)=x^2+1$ - By chain rule, $f^{\prime}(x)=3(g(x))^2 \cdot g^{\prime}(x)$ - $\begin{aligned}& =3\left(x^2+1\right)^2 \cdot 2 x \\& =6 x\left(x^2+1\right)^2\end{aligned}$ - or, $\quad y=\left(x^2+1\right)^3=u^3 ; u=x^2+1$ - $\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x} =3 u^2 \cdot 2 x$ - $=3\left(x^2+1\right)^2 \cdot 2 x$ - $=6 x\left(x^2+1\right)^2$ - $\rightarrow$ Expand $\left(x^2+1\right)^3=x^6+3 x^4+3 x^2+1$ - $\therefore f^{\prime}(x)=6 x^5+12 x^3+6 x$ </div> </main> <hr> <footer style = "font-size: 18px">Other forms of chain rule $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 1 $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Example 1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - $\rightarrow$ Use product rule: $f(x)=\left(x^2+1\right)\left(x^2+1\right)^2$ - $f^{\prime}(x) =2 x\left(x^2+1\right)^2+\left(x^2+1\right) \cdot \frac{d}{d x}\left(x^2+1\right)^2$ - $\frac{d}{d x}\left(x^2+1\right)^2 =2 x\left(x^2+1\right)+\left(x^2+1\right) \cdot 2 x$ - $=4 x\left(x^2+1\right)$ - $\therefore f^{\prime}(x) =2 x\left(x^2+1\right)^2+4 x\left(x^2+1\right)^2$ - $=6 x\left(x^2+1\right)^2$ - $=6 x\left(x^4+2 x^2+1\right)$ </div> <div style='float: right; width:1%;'> <!--   --> </div> </main> <hr> <footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 2 $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Example 2 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Find $\frac{d}{d x}\left[\sin \left(x^2\right)\right]$ - $y =\sin u, u=x^2$ - $\frac{d y}{d u} =\cos u, \frac{d u}{d x}=2 x$ - $\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=\cos \left(x^2\right) \cdot 2 x$ </div> </main> <hr> <footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 3 $\rightarrow$ Example 4 </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Example 3 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - Find $\frac{d}{d x}\left[\sin ^2\left(x^3\right)\right]$ - $y=\sin ^2\left(x^3\right)=\left[\sin \left(x^3\right)\right]^2$ - $\frac{d y}{d x}=2 \sin \left(x^3\right) \cdot \frac{d}{d x}\left[\sin \left(x^3\right)\right]=2 \sin \left(x^3\right) \cdot \cos \left(x^3\right) \cdot 3 x^2$ </div> </main> <hr> <footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 2 $\rightarrow$ <span style = "color:blue"> Example 3 </span> $\rightarrow$ Example 4 $\rightarrow$ Derivatives of inverse functions </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Example 4 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - $y= \sin \left(\sin \left(\cos \left(x^3+x\right)\right)\right)$ - $\frac{d y}{d x}= \cos \left(\sin \left(\cos \left(x^3+x\right)\right)\right) \cdot \frac{d}{d x}\left[\sin \left(\cos \left(x^3+x\right)\right)\right]$ - = $\cos \left(\sin \left(\cos \left(x^3+x\right)\right)\right) \cdot \cos \left[\cos \left(x^3+x\right)\right].{(-\sin (x^2+x))} \cdot(3 x^2+1)$ </div> <div style='float: right; width:1%;'> <!--   --> </div> </main> <hr> <footer style = "font-size: 18px">Example 2 $\rightarrow$ Example 3 $\rightarrow$ <span style = "color:blue"> Example 4 </span> $\rightarrow$ Derivatives of inverse functions $\rightarrow$ Derivatives of inverse functions </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Derivatives of inverse functions </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - $\frac{d}{d x}\left(\sin ^{-1} x\right), \frac{d}{d x}\left(\cos ^{-1} x\right), \frac{d}{d x}\left(\tan ^{-1} x\right), \cdots$ - Let $y=f(x)$ - Suppose $f(x)$ has an inverse $g(x)$. - ie. $\quad f(g(x))=g(f(x))=x$. - $y=f(x) \Leftrightarrow x =f^{-1}(y)=g(y) .$ - $\therefore \quad g^{\prime}(y) =\frac{d x}{d y}$ - $x=f(g(x)) \Rightarrow 1 =\frac{d}{d x}[f(g(x))]=f^{\prime}(g(x)) \cdot g^{\prime}(x)$ - So, if $f^{\prime}(g(x)) \neq 0$ then $g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Example 3 $\rightarrow$ Example 4 $\rightarrow$ <span style = "color:blue"> Derivatives of inverse functions </span> $\rightarrow$ Derivatives of inverse functions $\rightarrow$ Derivatives of inverse functions </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Derivatives of inverse functions </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - If $f^{-1}$ denotes the inverse of $f$, then - $\left.\frac{d}{d x}\left[f^{-1}(x)\right]=\frac{1}{f^{\prime}\left(f^{-1}(x)\right.}\right] \cdot f^{\prime}\left(f^{\prime}(x)\right) \neq 0 .$ - eg. $\frac{d}{d x}\left[\sin ^{-1} x\right]=\frac{1}{f^{\prime}\left(\sin ^{-1} x\right)}$, where $f(x)=\sin x$ - $=\frac{1}{\cos \left(\sin ^{-1} x\right)}$ - If $y=\sin ^{-1} x ; x=\sin y$ - $\Rightarrow \cos ^2 y=1-\sin ^2 y=1-x^2$ - $\Rightarrow \cos y= \pm \sqrt{1-x^2} .$ - $\sin ^{-1} x$ is defined for $x \in[-1,1]$ - $\sin ^{-1} x \in\left(0, \frac{\pi}{2}\right)$ if $x \in(0,1)$ - $\sin ^{-1} x \in\left(-\frac{\pi}{2}, 0\right)$ if $x \in(-1,0)$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Example 4 $\rightarrow$ Derivatives of inverse functions $\rightarrow$ <span style = "color:blue"> Derivatives of inverse functions </span> $\rightarrow$ Derivatives of inverse functions $\rightarrow$ Example 5 </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Derivatives of inverse functions </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - But $\cos \theta$ is positive if $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. - $\therefore \quad \cos \left(\sin ^{-1} x\right)>0 \quad \forall x \in(-1,1) .$ - $\therefore \cos \left(\sin ^{-1} x\right)=\sqrt{1-x^2} \quad \forall x \in(-1,1)$ - $\therefore \frac{d}{d x}\left[\sin ^{-1} x\right]=\frac{1}{\sqrt{1-x^2}}$ <!--If $f$ denotes the inverse of $f$, then--> <!--$\frac{d}{d x}\left[f^{-1}(x)\right]=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)} \cdot f^{\prime}\left(f^{\prime}(x)\right) \neq 0 \text {. }$ --> <!--eg.--> <!--$\frac{d}{d x}\left[\sin ^{-1} x\right] =\frac{1}{f^{\prime}\left(\sin ^{-1} x\right)}$ --> <!--where $f(x)=\sin x$--> <!--$ =\frac{1}{\cos \left(\sin ^{-1} x\right)}$--> <!--If--> <!--$y=\sin ^{-1} x, \left(sin y\right)$ --> <!--$\Rightarrow \cos ^2 y =1-\sin ^2 y=1-x^2$ --> <!--$\Rightarrow \cos y = \pm \sqrt{1-x^2} $ --> </div> <div style='float: right; width:1%;'> <!--   --> </div> </main> <hr> <footer style = "font-size: 18px">Derivatives of inverse functions $\rightarrow$ Derivatives of inverse functions $\rightarrow$ <span style = "color:blue"> Derivatives of inverse functions </span> $\rightarrow$ Example 5 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Example 5 </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> - $ \frac{d}{d x}\left[\tan ^{-1} x\right]=?$ - $y=\tan ^{-1} x \Leftrightarrow x=\tan y$ - $\quad \therefore \frac{d x}{d y}=\sec ^2 y$ - $\therefore \frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}=\frac{1}{\sec ^2 y}=\frac{1}{1+\tan ^2 y}=\frac{1}{1+x^2}$ - $\therefore \frac{d}{d x}\left[\tan ^{-1} x\right]=\frac{1}{1+x^2}$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Derivatives of inverse functions $\rightarrow$ Derivatives of inverse functions $\rightarrow$ <span style = "color:blue"> Example 5 </span> $\rightarrow$ Thank you $\rightarrow$ </footer>

### <Success style="font-size:25px"> Derivatives L-2 </Success> ### <primary style="font-size:24px"> Thank you </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;'> </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Derivatives of inverse functions $\rightarrow$ Example 5 $\rightarrow$ <span style = "color:blue"> Thank you </span> $\rightarrow$ $\rightarrow$ </footer>