Engineering Mathematics online quiz focuses on “Limits and Derivatives of Several Variables – 4”.
1. (frac{d(uvw)}{dx}) is where u ,v, w are the functions of x
a) u’vw + uv’w + uvw’
b) uvw + uv’w’ + u’v’w’
c) u’v’w + uv’w’ + u’vw’
d) uv’w’ + u’v’w’ + uvw
Answer: a
Explanation:
(frac{d(uvw)}{dx}=ufrac{d(uvw)}{dx}+vwfrac{d(u)}{dx}=u[vfrac{d(w)}{dx}+wfrac{d(v)}{dx}]+vwfrac{d(u)}{dx})=u’vw+uv’w+uvw’
2. (frac{d(frac{u}{v})}{dx}) is where u, v are the functions of x
a) v’u’ – uv ⁄ v2
b) vu’ – uv’ ⁄ v2
c) vu – u’v’ ⁄ v2
d) 0
Answer: b
Explanation:
(frac{d(frac{u}{v})}{dx}=ufrac{d(frac{1}{v})}{dx}+frac{1}{v}frac{d(u)}{dx}=-frac{u}{v^2}frac{d(v)}{dx}+frac{1}{v}frac{d(u)}{dx}=frac{vu’-uv’}{v^2})
3. If (y=frac{sin(x)e^x}{cos^2(x)}), find dy⁄dx .
a) Sec2 (x) ex [1 + Tan(x)] + ex Tan(x)Sec(x)
b) Sec2 (x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
c) Sec2 (x) e2x [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
d) Sec(x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
Answer: c
Explanation: (y=frac{sin(x)e^x}{cos^2(x)}) = Tan(x)Sec(x) ex
dy⁄dx = Sec2 (x)Sec(x) ex + Sec2 (x)Tan(x) ex + ex Tan(x)Sec(x)
dy⁄dx = Sec2 (x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x).
4. Value of d⁄dx [(1 + xex}{1-Cos(x))].
a) (frac{(1-Sin(x))(1+x) e^x + Cos(x)(1+xe^x)}{[1-Cos (x)]^2})
b) (frac{(1-Cos(x))(1+x) e^x + Sin(x)(1+xe^x)}{[1-Cos (x)]^4})
c) (frac{(1-Cos(x))(1+x) e^x + Sin(x)(1+xe^x)}{[1-Cos (x)]^2})
d) (frac{(1-Cos(x))(1+x) e^x – Sin(x)(1+xe^x)}{[1-Cos (x)]^2})
Answer: c
Explanation:
(frac{d}{dx} (1+xe^x)/(1-Cos(x)))
(frac{d}{dx} (1+xe^x)/(1-Cos(x)) = frac{(1-Cos(x))(1+x) e^x + Sin(x)(1+xe^x)}{[1-Cos (x)]^2})
5. Find the derivative of Sin(x)Tan(x) w.r.t ex Tan(x)
a) (frac{Sin(x)(1+Sec^4 (x))}{e^x (1+Tan^2 (x)+Tan(x))})
b) (frac{Sin(x)(1+Sec^2 (x))}{e^x (1+Tan^4 (x)+Tan(x))})
c) (frac{Sin(x)(1+Sec^2 (x))}{e^x (1+Tan^2 (x)+Tan(x))})
d) (frac{Sin(x)(1+Sec^2 (x))}{e^x (2+Tan^2 (x)+Tan(x))})
Answer: c
Explanation:
(frac{dSin(x)Tan(x)}{de^x Tan(x)} = frac{dSin(x)Tan(x))}{dx} frac{dx}{de^x Tan(x)})
=(frac{dSin(x)Tan(x)}{dx} frac{1}{frac{de^x Tan(x)}{dx}})
=(frac{[Sin(x)Sec^2 (x)+Sin(x)]}{e^x (Sec^2 (x)+Tan(x))})
=(frac{Sin(x)(1+Sec^2 (x))}{e^x (1+Tan^2 (x)+Tan(x))})
6. Evaluate (frac{d[Tan^n (x)+Tanx^n+Tan^{-1} x+Tan(nx)}{dx}]) is
a) (nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1+x^2)+nTan(nx)Sec^2 (nx))
b) (nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1+x^2)+nSec^2 (nx))
c) (nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1-x^2)+nSec^2 (nx))
d) (2nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1+x^2)+nSec^2 (nx))
Answer: b
Explanation: (frac{d[Tan^n (x)+Tanx^n+Tan^{-1} x+Tan(nx)}{dx}])
(=nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+frac{1}{1+x^2}+nSec^2 (nx)).
7. Evaluate d/dx xx ln(x)
a) x(x-1) + x2x ln(x) + xx [ln(x)]2
b) x(x-1) + xx ln(x) + xx [ln(x)]2
c) x(x-1) + xx ln(x) + xx ln(x)
d) xx + xx ln(x) + xx [ln(x)]2
Answer: b
Explanation:
(frac{dx^x ln(x)}{dx}=frac{x^x dln(x)}{dx}+frac{ln(x)dx^x}{dx})
Now
Y = xx
Taking log on both side.
ln(y) = xln(x)
Differntiating both sides
(frac{1}{y} frac{dy}{dx}=1+ln(x))
(frac{dy}{dx})=xx (1+ln(x))
Hence,
(frac{dx^x ln(x)}{dx}=x^{x-1}+x^x ln(x) (1+ln(x) )=x^{x-1}+x^x ln(x)+x^x [ln(x)]^2)
8. Evaluate the differentiation of (tan^{-1}frac{cos(x)-sin(x)}{cos(x)+sin(x)})
a) tan-1x
b) 1
c) 0
d) -1
Answer: b
Explanation:
(tan^{-1}frac{cos(x)-sin(x)}{cos(x)+sin(x)}=tan^{-1}tan(x)=x)
Hence
(frac{d}{dx}tan^{-1}frac{cos(x)-sin(x)}{cos(x)+sin(x)}=1)
9. If y = Tan(x)Tan(x) then dy⁄dx = ?
a) Tan(x) [1 + lnTan(x)] Tan(x)Tan(x)
b) Tan2 (x) [1 + lnTan(x)] Tan(x)Tan(x)
c) Sec2 (x) [1 + lnTan(x)] Tan(x)Tan(x)
d) Sec(x) [1 + lnTan(x)] Tan(x)Tan(x)
Answer: c
Explanation: y = Tan(x)Tan(x)
Taking ln on both side
ln y = Tan(x)lnTan(x)
Differentiating w.r.t x
(frac{1}{y} frac{dy}{dx} = frac{Tan(x)Sec^2 (x)}{Tan(x)}+Sec^2 (x)lnTan(x))
(frac{1}{y} frac{dy}{dx} = Sec^2 (x)[1+lnTan(x)])
(frac{dy}{dx}=Sec^2 (x)[1+lnTan(x)]Tan(x)^{Tan(x)})
10. Evaluate d⁄dx Cot(x)Cosec(x)
a) -Cosec2 (x) – Cosec2 (x)Cot(x)
b) -Cosec3 (x) – Cosec2 (x)Cot(x)
c) -Cosec(x) – Cosec2 (x)Cot(x)
d) -Cosec3 (x) – Cosec(x)Cot2 (x)
Answer: b
Explanation: d⁄dx Cot(x)Cosec(x) = -Cosec3 (x) – Cosec2 (x)Cot(x).(By multiplication rule)
11. Evaluate differentiation of x2 Sin(x) w.r.t Tan(x)Cosec(x)
a) (frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)-Sec^2 (x)Cosec(x)})
b) (frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)+Cos(x)Sin(x)})
c) (frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)+Sec^2 (x)Cosec(x)})
d) (frac{[2xSin(x)+x^2 Cos(x)]}{+Cosec(x)+Sec^2 (x)Cosec(x)})
Answer: c
Explanation:
(frac{dx^2 Sin(x)}{dTan(x)Csoec(x)}=frac{dx^2 Sin(x)}{dx} frac{dx}{dTan(x)Cosec(x)})
(=frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)+Sec^2 (x)Cosec(x)})
12. If z = ex Sin(Cos(x))Cos(Sin(x)) Then find dz⁄dx
a) [exSin(Cos(x))Cos(Sin(x))-exCos(x)Cos(Cos(x))Cos(Sin(x))-exSin(x)Sin(Cos(x))Sin(Sin(x))]
b) [exSin(Cos(x))Cos(Sin(x))-exSin(x)Cos(Cos(x))Cos(Sin(x))-exCos(x)Sin(Cos(x))Sin(Sin(x))]
c) [exCos(Cos(x))Sin(Sin(x))-exSin(x)Cos(Cos(x))Cos(Sin(x))-exCos(x)Sin(Cos(x))Sin(Sin(x))]
d) [exSin(Cos(x))Cos(Sin(x))-exCos(x)Cos(Cos(x))Cos(Sin(x))-exSin(x)Sin(Cos(x))Sin(Sin(x))]