Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

**Q2.**If y=cos^(-1)((5cosx-12sinx)/13) where x∈(0,Ï€/2), then dy/dx is

Solution

(a) Let cosa=5/(13,) then sina=12/13 So, y=cos^(-1)(cosa.cosx-sina.sinx) ⇒y=cos^(-1){cos(x+a)}=x+a (∵x+a is in the first or the second quadrant) ⇒dy/dx=1

(a) Let cosa=5/(13,) then sina=12/13 So, y=cos^(-1)(cosa.cosx-sina.sinx) ⇒y=cos^(-1){cos(x+a)}=x+a (∵x+a is in the first or the second quadrant) ⇒dy/dx=1

**Q4.**The nth derivative of the function f(x)=1/(1-x^2 ) (where x∈(-1,1)) at the point x=0 where n is even is

Solution

(b) f(x)=1+x^2+x^4+x^6+⋯∞, where |x|≤1 ⇒f^n (0)=n!, where n is even

(b) f(x)=1+x^2+x^4+x^6+⋯∞, where |x|≤1 ⇒f^n (0)=n!, where n is even

**Q5.**If y=√((1-x)/(1+x),) then (1-x^2 ) dy/dx is equal to

Solution

(c) We have y=√((1-x)/(1+x)) Differentiating w.r.t. x, we get dy/dx=1/2 ((1-x)/(1+x))^(1/2-1) d/dx ((1-x)/(1+x)) =1/2 √((1+x)/(1-x) )×((1+x)(-1)-(1-x)(1))/(1+x)^2 =-√((1+x)/(1-x)) 1/(1+x)^2 ⇒(1-x^2 ) dy/dx=-√((1+x)/(1-x)) 1/(1+x)^2 (1-x^2) ⇒(1-x)^2 dy/dx=-√((1-x)/(1+x) ) ⇒(1-x)^2 dy/dx=-y ⇒(1-x^2 ) dy/dx+y=0

(c) We have y=√((1-x)/(1+x)) Differentiating w.r.t. x, we get dy/dx=1/2 ((1-x)/(1+x))^(1/2-1) d/dx ((1-x)/(1+x)) =1/2 √((1+x)/(1-x) )×((1+x)(-1)-(1-x)(1))/(1+x)^2 =-√((1+x)/(1-x)) 1/(1+x)^2 ⇒(1-x^2 ) dy/dx=-√((1+x)/(1-x)) 1/(1+x)^2 (1-x^2) ⇒(1-x)^2 dy/dx=-√((1-x)/(1+x) ) ⇒(1-x)^2 dy/dx=-y ⇒(1-x^2 ) dy/dx+y=0

**Q7.**if f(x)=x+tanx and f is inverse of g, then g’(x) equals

Solution

(c) f(x)=x+tanx f(f^(-1)(y))=f^(-1) (y)+tanf^(-1)(y) y=g(y)+tang(y) x=g(x)+tang(x) Differentiating, we get 1=g'(x)+sec^2g(x)g'(x) ⇒g'(x)=1/(1+sec^2g(x))=1/(2+[g(x)-x]^2 )

(c) f(x)=x+tanx f(f^(-1)(y))=f^(-1) (y)+tanf^(-1)(y) y=g(y)+tang(y) x=g(x)+tang(x) Differentiating, we get 1=g'(x)+sec^2g(x)g'(x) ⇒g'(x)=1/(1+sec^2g(x))=1/(2+[g(x)-x]^2 )

**Q8.**If g is the inverse function of f and f' (x)=sinx, then g'(x) is

Solution

(a) Since g is the inverse function of f, we have f{g(x)}=x ⇒d/dx (f{g(x)})=1 ⇒f'{g(x)}.g'(x)=1 ⇒sin{g(x)} g'(x)=1 ⇒g'(x)=1/sin{g(x)}

(a) Since g is the inverse function of f, we have f{g(x)}=x ⇒d/dx (f{g(x)})=1 ⇒f'{g(x)}.g'(x)=1 ⇒sin{g(x)} g'(x)=1 ⇒g'(x)=1/sin{g(x)}