Question
Differentiate the following w.r.t. x:
$\sin\sqrt{\text{x}}+\cos^2\sqrt{\text{x}}$
$\sin\sqrt{\text{x}}+\cos^2\sqrt{\text{x}}$
✓
Answer
Let $\text{y}=\sin\sqrt{\text{x}}+\big(\cos^2\sqrt{\text{x}}\big)^2$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin\Big(\text{x}^{\frac{1}{2}}\Big)+\frac{\text{d}}{\text{dx}}\Big[\cos\Big(\text{x}^{\frac{1}{2}}\Big)\Big]^2$
$=\cos\text{x}^{\frac{1}{2}}\cdot\frac{\text{d}}{\text{dx}}\text{x}^{\frac{1}{2}}+\cos\Big(\text{x}^{\frac{1}{2}}\Big)\frac{\text{d}}{\text{dx}}\Big[\cos\Big(\text{x}^{\frac{1}{2}}\Big)\Big]$
$=\cos\sqrt{\text{x}}\cdot\frac{1}{2\sqrt{\text{x}}}+2\cos\sqrt{\text{x}}\Big[-\sin\sqrt{\text{x}}\cdot\frac{1}{2\sqrt{\text{x}}}\Big]$
$=\frac{1}{2\sqrt{\text{x}}}\Big[\cos\big(\sqrt{\text{x}}\big)-\sin\big(2\sqrt{\text{x}}\big)\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin\Big(\text{x}^{\frac{1}{2}}\Big)+\frac{\text{d}}{\text{dx}}\Big[\cos\Big(\text{x}^{\frac{1}{2}}\Big)\Big]^2$
$=\cos\text{x}^{\frac{1}{2}}\cdot\frac{\text{d}}{\text{dx}}\text{x}^{\frac{1}{2}}+\cos\Big(\text{x}^{\frac{1}{2}}\Big)\frac{\text{d}}{\text{dx}}\Big[\cos\Big(\text{x}^{\frac{1}{2}}\Big)\Big]$
$=\cos\sqrt{\text{x}}\cdot\frac{1}{2\sqrt{\text{x}}}+2\cos\sqrt{\text{x}}\Big[-\sin\sqrt{\text{x}}\cdot\frac{1}{2\sqrt{\text{x}}}\Big]$
$=\frac{1}{2\sqrt{\text{x}}}\Big[\cos\big(\sqrt{\text{x}}\big)-\sin\big(2\sqrt{\text{x}}\big)\Big]$
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