Q.Differentiate $y = 7x^{5} + 4\sqrt{x} - \dfrac{3}{x}$.

With $f(x)=mx+c$, $f(0)=c=1$ and $f'(0)=m=1$, so $f(x)=x+1$ and $f(2)=3$.

For the linear function $f(x)=mx+c$:

$f(0)=m(0)+c=c$. Given $f(0)=1$, so $c=1$.

$f'(x)=m$ (the derivative of a linear function is its constant slope). Given $f'(0)=1$, so $m=1$.

Hence $f(x)=x+1$.

$f(2)=2+1=3$.

Since $7>5$, $f(x)=x-5$ near $x=7$, so $f'(x)=1$ there, giving $f'(7)=1$.

$f(x)=|x-5|=\begin{cases}x-5 & x\ge5\\ 5-x & x<5\end{cases}$

At $x=7$ (which is $>5$), $f(x)=x-5$ in a neighborhood of $x=7$, so $f$ is differentiable there with $f'(x)=1$.

$f'(7)=1$.

Near $x=-3$ (which is negative), $f(x)=x|x|$ simplifies to $-x^2$, whose derivative is $-2x$.

Since $x=-3<0$, in a neighbourhood of $-3$ we have $|x|=-x$, so $f(x)=x|x|=x(-x)=-x^2$.

Differentiating, $f'(x)=-2x$.

At $x=-3$: $f'(-3)=-2(-3)=6$.

Since calculus derivatives of sine require the angle in radians, first rewrite $x^\circ$ as $\dfrac{\pi x}{180}$ radians.

We have $\sin x^\circ=\sin\left(\dfrac{\pi x}{180}\right)$. So

$\dfrac{d}{dx}\left(\dfrac{2}{\pi}\sin x^\circ\right)=\dfrac{2}{\pi}\cdot\cos\left(\dfrac{\pi x}{180}\right)\cdot\dfrac{\pi}{180}=\dfrac{2}{180}\cos x^\circ=\dfrac{1}{90}\cos x^\circ.$

Applying the power rule, $\dfrac{d}{dx}(ax^n) = anx^{n-1}$ — option (b).

GSEB Class-12 Statistics, Differentiation chapter:

The power rule states $\dfrac{d}{dx}(x^n) = n x^{n-1}$, and a constant multiplier is carried through:

$\frac{dy}{dx} = \frac{d}{dx}(a x^n) = a \cdot n x^{n-1} = an x^{n-1}$

With $f(x)=mx+c$, the condition $f(0)=1$ gives $c=1$ and $f'(0)=1$ gives $m=1$, so $f(3)=4$.

$f(x)=mx+c \Rightarrow f(0)=c$. Given $f(0)=1$, so $c=1$.

$f'(x)=m$ for all $x$ (the derivative of a linear function is its slope, constant everywhere). Given $f'(0)=1$, so $m=1$.

Thus $f(x)=x+1$, and $f(3)=3+1=4$.

Solving $x^2-3x = 2x-3$ leads to $x^2-5x+3=0$, whose discriminant $13$ is not a perfect square, so both roots $\dfrac{5\pm\sqrt{13}}{2}$ are irrational.

$f(x)=x^2-3x$, so $f'(x)=2x-3$.

Set $f(x)=f'(x)$: $x^2-3x = 2x-3$.

Rearrange: $x^2-3x-2x+3 = 0 \Rightarrow x^2-5x+3=0$.

Discriminant $= (-5)^2-4(1)(3) = 25-12 = 13$, which is not a perfect square.

Roots: $x = \dfrac{5\pm\sqrt{13}}{2}$ — since $\sqrt{13}$ is irrational, both roots are irrational.