Question 28.7.1: Differentiate (i) sin x³ (ii) sin³ x (iii) e^sin x...
Differentiate (i) \sin x^{3} \qquad(ii) \sin ^{3} x \qquad(iii) e^{\sin x}
(i) Let y=\sin x^{3} .
Put x^{3}=t , so that y=\sin t and t=x^{3} .
\therefore \quad \frac{d y}{d t}=\cos t and \frac{d t}{d x}=3 x^{2} .
So, \quad \frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)\\ \\=3 x^{2} \cos t=3 x^{2} \cos x^{3} \quad\left[\because t=x^{3}\right] .
Hence, \frac{d}{d x}\left(\sin x^{3}\right)=3 x^{2} \cos x^{3} .
(ii) Let y=\sin ^{3} x=(\sin x)^{3} .
Put \sin x=t , so that y=t^{3} and t=\sin x .
\therefore \quad \frac{d y}{d t}=3 t^{2} and \frac{d t}{d x}=\cos x .
Hence, \frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)\\ =3 t^{2} \cos x=3 \sin ^{2} x \cos x \quad[\because t=\sin x] .
(iii) Let y=e^{\sin x} .
Put \sin x=t , so that y=e^{t} and t=\sin x \\ \therefore \quad \frac{d y}{d t}=e^{t} and \frac{d t}{d x}=\cos x .
Hence, \frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right) \\=e^{t} \cos x=e^{\sin x} \cdot \cos x \quad[\because t=\sin x] .