Applications of Derivatives

y (x) = 2^x – x²

y? (x) = 2x log 2 – 2x

M = 3

N = 2

M + N = 5

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$

Equation of tangent y – t³ = 3t² (x – t) 

Let again meet the curve at $Q\left(t_1,t_1^3\right)$

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$

$t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$

$t_1^2+t{t}_1-2t^2=0$

=> t₁ = -2t

Required ordinate = $\frac{2t^3+t_1^3}{3}=\frac{2t^3-8t^3}{3}=-2t^3$   

Given f(X) = ${\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{\left(1+t\right)}dt............\left(i\right)}$  

So   $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^{1/x}\frac{\lambda lo{g}_{e}t}{1+t}dt}............\left(ii\right)$

put $t=\frac{1}{z}thendt=-\frac{1}{z^2}dz$  

$f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{t\left(1+t\right)}}dt............\left(iii\right)$

(i) + (iii), f(x) + $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\left(\frac{lo{g}_{e}t}{1+t}+\frac{lo{g}_{e}t}{t\left(1+t\right)}\right)dt}$

$={\left[\frac{{\left(lo{g}_{e}t\right)}^2}{2}\right]}_1^x=\frac{{\left(lo{g}_{x}x\right)}^2}{2}$

Hence f(e) + $f\left(\frac{1}{e}\right)=\frac{1}{2}$

$\frac{dy}{dx}=3a{x}^2-2bx$

${\frac{dy}{dx}|}_{x=1}=3a-2b=3$

$a=\frac{2b+3}{3}\in \left[1,2\right]$

$2b+3\in \left[3,6\right]$

$2b\in \left[0,3\right]$

$b\in \left[0,\frac{3}{2}\right]$

f' (x) = cosx + sinx − k ≤ 0∀x ∈ R
k ≥ √2

f (x) = x? /20 - x? /12 + 5
f' (x) = x? /4 - x³/3 = x³ (x/4 - 1/3)
Local maxima at 0, Local minima at 4/3
f' (x) = x³ - x² = x² (x-1)
x = 1 point of inflection

y = (3/2)sin (2θ)
x = e^θ sinθ
dy/dθ = 3cos (2θ)
dx/dθ = e^θ (cosθ + sinθ)
dy/dx = (3cos (2θ) / (e^θ (cosθ + sinθ) = (3 (cosθ - sinθ) / e^θ

f' (0) = 0   

f (x) = (4a - 3) (x + ln5) + 2 (a - 7)cot (x/2)sin² (x/2)
= (4a - 3) (x + ln5) + (a - 7)sin (x)
f' (x) = (4a - 3) + (a - 7)cos (x)
For critical points f' (x) = 0
cos (x) = - (4a - 3) / (a - 7) = (3 - 4a) / (a - 7)
⇒ -1 ≤ (3 - 4a) / (a - 7) ≤ 1
Solving this inequality leads to:
⇒ a ∈ [4/3, 2]

f (2)=8, f' (2)=5, f' (x) ≥ 1, f' (x) ≥ 4, ∀x ∈ (1,6)
Using LMVT
f' (x) = (f' (5) - f' (2)/ (5-2) ≥ 4 ⇒ f' (5) ≥ 17
f' (x) = (f (5) - f (2)/ (5-2) ≥ 1 ⇒ f (5) ≥ 11
Therefore f' (5) + f (5) ≥ 28