NCERT Solutions · Class 12 Mathematics Application of Derivatives

Use the first-derivative test for monotonicity: if \(\displaystyle f'(x)>0\) at every point of an interval, then \(\displaystyle f\) is strictly increasing on that interval.Here \(\displaystyle f(x)=3x+17\), so \[f'(x)=\frac{d}{dx}\left(3x+17\right)=3 .\]Thus \(\displaystyle f'(x)=3>0\) for every \(\displaystyle x\in\mathbf{R}\).Hence \(\displaystyle f(x)=3x+17\) is strictly increasing on \(\displaystyle \mathbf{R}\).

Apply the test: \(\displaystyle f'(x)>0\) on an interval \(\displaystyle \Rightarrow\) \(\displaystyle f\) is strictly increasing there.With \(\displaystyle f(x)=e^{2x}\), the chain rule gives \[f'(x)=e^{2x}\cdot\frac{d}{dx}(2x)=2e^{2x}. \]The exponential function is positive for every real argument, so \(\displaystyle e^{2x}>0\) for all \(\displaystyle x\in\mathbf{R}\), and therefore \[f'(x)=2e^{2x}>0 \quad\text{for all } x\in\mathbf{R}. \]Hence \(\displaystyle f(x)=e^{2x}\) is strictly increasing on \(\displaystyle \mathbf{R}\).

\[y=\log(1+x)-\frac{2x}{2+x},\qquad x>-1. \]Differentiate, using the quotient rule on the second term: \[\frac{dy}{dx}=\frac{1}{1+x}-\frac{2(2+x)-2x(1)}{(2+x)^{2}}=\frac{1}{1+x}-\frac{4}{(2+x)^{2}}. \]Put the two terms over the common denominator \(\displaystyle (1+x)(2+x)^{2}\): \[\frac{dy}{dx}=\frac{(2+x)^{2}-4(1+x)}{(1+x)(2+x)^{2}}=\frac{4+4x+x^{2}-4-4x}{(1+x)(2+x)^{2}}=\frac{x^{2}}{(1+x)(2+x)^{2}}. \]Now use the domain restriction \(\displaystyle x>-1\): it gives \(\displaystyle 1+x>0\), and also \(\displaystyle 2+x>1>0\) so \(\displaystyle (2+x)^{2}>0\). The numerator \(\displaystyle x^{2}\ge 0\). Hence \[\frac{dy}{dx}\ge 0 \quad\text{for all } x>-1, \] with equality only at the single point \(\displaystyle x=0\).Since the derivative is non-negative throughout the domain and vanishes only at an isolated point, \(\displaystyle y\) is an increasing function of \(\displaystyle x\) for all \(\displaystyle x>-1\).

\[y=[x(x-2)]^{2}=\left(x^{2}-2x\right)^{2}. \]By the chain rule, \[\frac{dy}{dx}=2\left(x^{2}-2x\right)(2x-2)=2\cdot x(x-2)\cdot 2(x-1)=4x(x-1)(x-2). \]The derivative vanishes at \(\displaystyle x=0,1,2\), which split \(\displaystyle \mathbf{R}\) into four intervals. Test one point in each:\(\displaystyle x=-1:\ 4(-1)(-2)(-3)=-24<0\).\(\displaystyle x=\tfrac12:\ 4\left(\tfrac12\right)\left(-\tfrac12\right)\left(-\tfrac32\right)=\tfrac32>0\).\(\displaystyle x=\tfrac32:\ 4\left(\tfrac32\right)\left(\tfrac12\right)\left(-\tfrac12\right)=-\tfrac32<0\).\(\displaystyle x=3:\ 4(3)(2)(1)=24>0\).So \(\displaystyle \dfrac{dy}{dx}>0\) on \(\displaystyle (0,1)\) and on \(\displaystyle (2,\infty)\).Hence \(\displaystyle y\) is an increasing function for \(\displaystyle 0\le x\le 1\) and for \(\displaystyle x\ge 2\).

\[y=\frac{4\sin\theta}{2+\cos\theta}-\theta. \]Differentiate the first term by the quotient rule: \[\frac{dy}{d\theta}=\frac{4\cos\theta(2+\cos\theta)-4\sin\theta(-\sin\theta)}{(2+\cos\theta)^{2}}-1=\frac{8\cos\theta+4\cos^{2}\theta+4\sin^{2}\theta}{(2+\cos\theta)^{2}}-1. \]Use \(\displaystyle \sin^{2}\theta+\cos^{2}\theta=1\) in the numerator: \[\frac{dy}{d\theta}=\frac{8\cos\theta+4}{(2+\cos\theta)^{2}}-1=\frac{8\cos\theta+4-\left(4+4\cos\theta+\cos^{2}\theta\right)}{(2+\cos\theta)^{2}}=\frac{4\cos\theta-\cos^{2}\theta}{(2+\cos\theta)^{2}}. \]Factor the numerator: \[\frac{dy}{d\theta}=\frac{\cos\theta\,(4-\cos\theta)}{(2+\cos\theta)^{2}}. \]For \(\displaystyle 0<\theta<\frac{\pi}{2}\): \(\displaystyle \cos\theta>0\); and since \(\displaystyle |\cos\theta|\le 1\), \(\displaystyle 4-\cos\theta\ge 3>0\); the denominator \(\displaystyle (2+\cos\theta)^{2}>0\). Therefore \[\frac{dy}{d\theta}>0 \quad\text{on }\left(0,\tfrac{\pi}{2}\right), \] and at the endpoint \(\displaystyle \theta=\frac{\pi}{2}\) it is \(\displaystyle 0\).Hence \(\displaystyle y\) is an increasing function of \(\displaystyle \theta\) on \(\displaystyle \left[0,\frac{\pi}{2}\right]\).

Let \(\displaystyle f(x)=\log x\), whose domain is \(\displaystyle (0,\infty)\).\[f'(x)=\frac{1}{x}. \]For every \(\displaystyle x\) in the domain, \(\displaystyle x>0\), and hence \[f'(x)=\frac{1}{x}>0 . \]Since the derivative is positive at every point of \(\displaystyle (0,\infty)\), the logarithmic function is strictly increasing on \(\displaystyle (0,\infty)\).