H.S.C.

Differentiation EX 1.1 Q.4 Q.5 Q.6 Q.7 Mathematics and statistic for arts and science

10:50 PM
DERIVATIVE 
EX 1.1 

 Q.4

 i) $$\begin{array}{rl}f(x)&=f[g(x)] \\ f^{\prime}(x) & =\frac{d}{d x} f(x) \\ f^{\prime}(x) & =\frac{d}{d x} f[g(x)] \\ & =\frac{d}{d x} f\left(g(x) \cdot \frac{d}{d x} g(x)\right. \\ f^{\prime}(x) & =f^{\prime} g(x) \cdot g^{\prime}(x) \\ Now,\\ {f}^{\prime}(2)&=f^{\prime} g(2) \cdot g^{\prime}(2)\\ &=f^{\prime}(6) \cdot 4 \quad\\ &=-4 \times 4\\ &=-16 \end{array}$$
ii) $$\begin{array}{rl} R(x)&=g[3+f(x)]\\ R^{\prime}(x) &=g^{\prime}(3+f(x)] \cdot \frac{d}{d x}[3+f(x)] \\ &=g^{\prime}[3+f(x)]\left[\frac{d}{d x}(3)+\frac{d}{d x} f(x)\right] \\ R^{\prime}(x) &=g^{\prime}[3+f(x)]\left[0+f^{\prime}(x)\right] \\ Now \\ R^{\prime}(4)&=g^{\prime}[3+f(4)]\left[f^{\prime}(4)\right]\\ &=g^{\prime}[3+3] \cdot 5\\ &=g^{\prime}(6) \cdot 5\\ R^{\prime}(4)&=35 \end{array}$$ 
iii) $$\begin{array}{rl} If\\ s(x)&=f(9-f(x)] \\ s^{\prime}(x)&=\frac{d}{d x} f(9-f(x)]\\ &=f^{\prime}[9-f(x)] \cdot \frac{d}{d x}(9-f(x)]\\ &=f^{\prime}(9-f(x)]\left[\frac{d}{d x} 9-\frac{d}{d x} f(x)\right]\\ s^{\prime}(x)&=f^{\prime}(9-f(x)]\left[0-f^{\prime}(x)\right]\\ Now,\\ s^{\prime}(4)&=f^{\prime}[9-t(4)] f^{\prime}(4)\\ &=f^{\prime}[9-3] \cdot 5\\ &=t^{\prime}(6) \cdot 5\\ S^{\prime}(4)&=-20 \end{array}$$
iv) $$\begin{array}{rl} If \\ S(x)&=g[g(x)] \\ S^{\prime}(x)&=\frac{d}{d x} g[g(x)]\\ &=g[g(x)] \cdot \frac{d}{d x} g(x)\\ S^{\prime}(x)&=g^{\prime}(g(x)] \cdot g^{\prime}(x)\\ Now, \\ S^{\prime}(6)&=g^{\prime}(g(6)] \cdot g^{\prime}(6)\\ &=g^{\prime}(2) \cdot 7\\ &=4\\ S^{\prime}(6)&=28 \end{array}$$

Q.5
$\begin{array}{rl} f^{\prime}(3)&=-1\\ g^{\prime}(2)&=5\\ g(2)&=3\\ y&=f[g(x)]\\ \frac{d y}{d x}&=\frac{d}{d x} f[g(x)]\\ &=f^{\prime} g(x) \cdot \frac{d}{d x} g(x) \\ &=f^{\prime} g(x) \cdot g^{\prime}(x)\\ Now,\\ \left[\frac{d y}{d x}\right]_{x=2}&=f^{\prime} g(2) \cdot g^{\prime}(2)\\ &=f^{\prime}(3) \cdot 5\\ &=-1.5\\ &\left[\frac{d y}{d x}\right]_{x=2}=-5 \end{array}$

Q.6
$\begin{array}{rl} \quad f(1)&=4\\ g(1)&=3\\ f^{\prime}(1)&=3\\ g^{\prime}(1)&=4\\ h(x)&=\sqrt{4 f(x)+3 g(x)}\\ h^{\prime}(x)&=\frac{d}{d x} \sqrt{f f(x)}+3 g(x) \\ &=\frac{1}{2 \sqrt{1 f(x)+3 g(x)}} \cdot \frac{d}{d x}(4 f(x)+3 g(x)]\\ h^{\prime}(x)&=\frac{1}{e \sqrt{4 f(x)+3 g(x)}} \cdot\left[9 \cdot f^{\prime}(x)+3 \cdot g^{\prime}(x)\right]\\ Now, \\ \quad h^{\prime}(1)&=\frac{1}{2 \sqrt{+f(1)+3 g(1)}} \cdot\left[f f^{\prime}(1)+3 g^{\prime}(1)\right]\\ &=\frac{1}{2 \sqrt{1.4+3.3}} \cdot[4 \cdot 3+3.4]\end{array}$






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